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Chapter 4. Stoichiometry of Chemic Reactions

four.1 Writing and Balancing Chemical Equations

Learning Objectives

By the finish of this section, y'all will be able to:

• Derive chemical equations from narrative descriptions of chemic reactions.
• Write and remainder chemical equations in molecular, full ionic, and cyberspace ionic formats.

The preceding chapter introduced the use of chemical element symbols to represent private atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately correspond these species. Extending this symbolism to stand for both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemic equation. Consider as an example the reaction between one methane molecule (CH₄) and two diatomic oxygen molecules (O₂) to produce one carbon dioxide molecule (CO_two) and two water molecules (H₂O). The chemic equation representing this procedure is provided in the upper half of Figure one, with space-filling molecular models shown in the lower one-half of the effigy.

Effigy 1. The reaction betwixt methane and oxygen to yield carbon dioxide and water (shown at bottom) may be represented by a chemical equation using formulas (top).

This example illustrates the fundamental aspects of any chemical equation:

1. The substances undergoing reaction are called reactants, and their formulas are placed on the left side of the equation.
2. The substances generated past the reaction are chosen products, and their formulas are placed on the right sight of the equation.
3. Plus signs (+) dissever private reactant and product formulas, and an pointer (⟶) separates the reactant and product (left and right) sides of the equation.
4. The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.

Information technology is common practice to use the smallest possible whole-number coefficients in a chemical equation, every bit is done in this example. Realize, notwithstanding, that these coefficients correspond the relative numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methyl hydride and oxygen react to yield carbon dioxide and water in a one:2:one:ii ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or two-4-2-iv, or 3-6-3-6, and so on (Figure 2). Likewise, these coefficients may exist interpreted with regard to whatever amount (number) unit, and so this equation may be correctly read in many ways, including:

• 1 methane molecule and two oxygen molecules react to yield ane carbon dioxide molecule and 2 h2o molecules.
• One dozen methane molecules and two dozen oxygen molecules react to yield one dozen carbon dioxide molecules and 2 dozen water molecules.
• One mole of methyl hydride molecules and ii moles of oxygen molecules react to yield one mole of carbon dioxide molecules and 2 moles of h2o molecules.

Figure ii. Regardless of the absolute numbers of molecules involved, the ratios betwixt numbers of molecules of each species that react (the reactants) and molecules of each species that course (the products) are the same and are given by the chemical reaction equation.

Balancing Equations

The chemical equation described in department four.1 is balanced, significant that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of affair. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparison these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of whatever formula containing that element by the chemical element's subscript in the formula. If an chemical element appears in more 1 formula on a given side of the equation, the number of atoms represented in each must be computed and so added together. For example, both product species in the case reaction, CO₂ and H₂O, comprise the element oxygen, and and then the number of oxygen atoms on the product side of the equation is

[latex](1 \;\text{CO}_2 \;\text{molecule} \times \frac{ii \;\text{O atoms}}{\text{CO}_2 \;\text{molecule}}) + (2\;\text{H}_2\text{O molecule} \times \frac{ane \;\text{O atom}}{\text{H}_2\text{O molecule}}) = 4 \;\text{O atoms}[/latex]

The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, every bit shown here:

[latex]\text{CH}_4 + two\text{O}_2 \longrightarrow \text{CO}_2 + ii\text{H}_2\text{O}[/latex]

Chemical element	Reactants	Products	Balanced?
C	1 × 1 = i	i × 1 = 1	1 = 1, yes
H	4 × 1 = 4	two × 2 = 4	four = 4, yes
O	two × 2 = 4	(ane × ii) + (ii × i) = iv	iv = four, yes
Table 1.

A balanced chemic equation oftentimes may be derived from a qualitative clarification of some chemical reaction by a adequately simple arroyo known every bit balancing past inspection. Consider as an instance the decomposition of h2o to yield molecular hydrogen and oxygen. This procedure is represented qualitatively by an unbalanced chemical equation:

[latex]\text{H}_2\text{O} \longrightarrow \text{H}_2 + \text{O}_2 \;(\text{unbalanced})[/latex]

Comparing the number of H and O atoms on either side of this equation confirms its imbalance:

Element	Reactants	Products	Balanced?
H	1 × 2 = 2	1 × two = 2	ii = ii, yes
O	1 × one = i	one × 2 = 2	ane ≠ two, no
Table two.

The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are non. To achieve residue, the coefficients of the equation may be changed as needed. Keep in mind, of class, that the formula subscripts define, in part, the identity of the substance, then these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H₂O to H₂O_two would yield balance in the number of atoms, just doing so also changes the reactant'southward identity (it'southward now hydrogen peroxide and non water). The O cantlet balance may exist achieved by changing the coefficient for H_iiO to 2.

[latex]2\text{H}_2\text{O} \longrightarrow \text{H}_2 + \text{O}_2 \;(\text{unbalanced})[/latex]

Element	Reactants	Products	Balanced?
H	2 × two = 4	one × 2 = 2	4 ≠ 2, no
O	2 × 1 = ii	ane × two = 2	ii = 2, aye
Table 3.

The H cantlet balance was upset by this modify, just it is hands reestablished by irresolute the coefficient for the H_ii product to 2.

[latex]ii\text{H}_2\text{O} \longrightarrow 2\text{H}_2 + \text{O}_2 \;(\text{counterbalanced})[/latex]

Element	Reactants	Products	Counterbalanced?
H	two × 2 = 4	2 × two = four	four = four, yep
O	2 × 1 = 2	1 × two = two	2 = 2, yep
Tabular array four.

These coefficients yield equal numbers of both H and O atoms on the reactant and production sides, and the counterbalanced equation is, therefore:

[latex]ii\text{H}_2\text{O} \longrightarrow 2\text{H}_2 + \text{O}_2[/latex]

Instance one

Balancing Chemic Equations
Write a balanced equation for the reaction of molecular nitrogen (North₂) and oxygen (O_ii) to form dinitrogen pentoxide.

Solution
First, write the unbalanced equation.

[latex]\text{Due north}_2 + \text{O}_2 \longrightarrow \text{N}_2 \text{O}_5 \;(\text{unbalanced})[/latex]

Next, count the number of each blazon of atom present in the unbalanced equation.

Chemical element	Reactants	Products	Balanced?
N	one × 2 = 2	1 × two = 2	2 = 2, yes
O	1 × ii = two	1 × 5 = five	ii ≠ five, no
Tabular array 5.

Though nitrogen is balanced, changes in coefficients are needed to residue the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O₂ and North₂O₅ to integers that will yield 10 O atoms (the to the lowest degree mutual multiple for the O cantlet subscripts in these two formulas).

[latex]\text{N}_2 + 5\text{O}_2 \longrightarrow 2\text{N}_2\text{O}_5 \;(\text{unbalanced})[/latex]

Chemical element	Reactants	Products	Balanced?
N	1 ×× two = 2	2 × ii = four	two ≠ four, no
O	5 × two = ten	two × 5 = ten	10 = x, yes
Tabular array 6.

The Northward atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N₂ to two.

[latex]2\text{N}_2 + 5\text{O}_2 \longrightarrow ii\text{N}_2 \text{O}_5[/latex]

Chemical element	Reactants	Products	Balanced?
N	ii × 2 = iv	2 × ii = 4	4 = 4, yeah
O	five × 2 = 10	2 × 5 = 10	10 = ten, yes
Table vii.

The numbers of Due north and O atoms on either side of the equation are now equal, and and so the equation is balanced.

Check Your Learning
Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen final, since it is present in more than one molecule on the correct side of the equation.)

Answer:

[latex]ii\text{NH}_4 \text{NO}_3 \longrightarrow 2\text{N}_2 + \text{O}_2 + 4\text{H}_2\text{O}[/latex]

It is sometimes user-friendly to apply fractions instead of integers as intermediate coefficients in the process of balancing a chemic equation. When balance is accomplished, all the equation's coefficients may then be multiplied by a whole number to catechumen the partial coefficients to integers without upsetting the atom balance. For instance, consider the reaction of ethane (C₂H₆) with oxygen to yield H₂O and CO₂, represented by the unbalanced equation:

[latex]\text{C}_2 \text{H}_6 + \text{O}_2 \longrightarrow \text{H}_2 \text{O} + \text{C} \text{O}_2 \;(\text{unbalanced})[/latex]

Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, equally shown:

[latex]\text{C}_2 \text{H}_6 + \text{O}_2 \longrightarrow 3\text{H}_2 \text{O} + ii\text{C} \text{O}_2 \;(\text{unbalanced})[/latex]

This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient tin can exist used with the O₂ reactant to yield an odd number, so a fractional coefficient, [latex]\frac{7}{2}[/latex], is used instead to yield a provisional balanced equation:

[latex]\text{C}_2 \text{H}_6 + \frac{7}{2}\text{O}_2 \longrightarrow three\text{H}_2 \text{O} + two\text{C} \text{O}_2 \;[/latex]

A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:

[latex]ii\text{C}_2 \text{H}_6 + 7\text{O}_2 \longrightarrow six\text{H}_2 \text{O} + 4\text{C} \text{O}_2 \;[/latex]

Finally with regard to balanced equations, retrieve that convention dictates employ of the smallest whole-number coefficients. Although the equation for the reaction betwixt molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, counterbalanced,

[latex]3\text{N}_2 + 9\text{H}_2 \longrightarrow 6\text{Northward} \text{H}_3[/latex]

the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient past the greatest common factor, iii, gives the preferred equation:

[latex]\text{N}_2 + three\text{H}_2 \longrightarrow ii\text{N} \text{H}_3[/latex]

Use this interactive tutorial for additional practice balancing equations.

Boosted Information in Chemical Equations

The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abridgement following the formulas. Common abbreviations include s for solids, l for liquids, g for gases, and aq for substances dissolved in water (aqueous solutions, as introduced in the preceding chapter). These notations are illustrated in the instance equation here:

[latex]2\text{Na}(s) + 2\text{H}_2 \text{O}(l) \longrightarrow 2\text{NaOH}(aq) + \text{H}_2(thousand)[/latex]

This equation represents the reaction that takes place when sodium metallic is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic chemical compound sodium hydroxide (a solid in pure form, but readily dissolved in h2o).

Special conditions necessary for a reaction are sometimes designated by writing a give-and-take or symbol to a higher place or below the equation's arrow. For example, a reaction carried out by heating may be indicated past the uppercase Greek alphabetic character delta (Δ) over the arrow.

[latex]\text{CaCO}_3(s) \;\xrightarrow{\Delta} \; \text{CaO}(s) + \text{CO}_2(k)[/latex]

Other examples of these special weather volition be encountered in more depth in later chapters.

Equations for Ionic Reactions

Given the affluence of water on earth, information technology stands to reason that a keen many chemic reactions accept identify in aqueous media. When ions are involved in these reactions, the chemical equations may be written with various levels of item appropriate to their intended use. To illustrate this, consider a reaction betwixt ionic compounds taking identify in an aqueous solution. When aqueous solutions of CaCl_two and AgNO₃ are mixed, a reaction takes identify producing aqueous Ca(NO₃)_two and solid AgCl:

[latex]\text{CaCl}_2(aq) + 2\text{AgNO}_3(aq) \longrightarrow \text{Ca(NO}_3)_2(aq) + two\text{AgCl}(s)[/latex]

This counterbalanced equation, derived in the usual fashion, is chosen a molecular equation because it doesn't explicitly represent the ionic species that are nowadays in solution. When ionic compounds dissolve in water, they may dissociate into their constituent ions, which are afterwards dispersed homogenously throughout the resulting solution (a thorough word of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more than realistically represented as dissociated ions, in this case:

[latex]\begin{array}{r @{{}\longrightarrow{}} l} \text{CaCl}_2(aq) & \text{Ca}^{two+}(aq) + two \text{Cl}^{-}(aq) \\[0.5em] 2 \text{AgNO}_3(aq) & two\text{Ag}^{+}(aq) + two {\text{NO}_3}^{-}(aq) \\[0.5em] \text{Ca(NO}_3)_2(aq) & \text{Ca}^{2+}(aq) + 2 {\text{NO}_3}^{-}(aq) \end{assortment}[/latex]

Unlike these three ionic compounds, AgCl does not dissolve in water to a significant extent, as signified by its physical state annotation, s.

Explicitly representing all dissolved ions results in a consummate ionic equation. In this particular case, the formulas for the dissolved ionic compounds are replaced past formulas for their dissociated ions:

[latex]\text{Ca}^{2+}(aq) + ii\text{Cl}^{-}(aq) + 2\text{Ag}^{+}(aq) + two{\text{NO}_3}^{-}(aq) \longrightarrow \text{Ca}^{2+}(aq) + 2{\text{NO}_3}^{-}(aq) + two\text{AgCl}(s)[/latex]

Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, Ca²⁺(aq) and NO3−(aq).NO3−(aq). These spectator ions—ions whose presence is required to maintain charge neutrality—are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a net ionic equation:

[latex]\rule[0.5ex]{4em}{0.1ex}\hspace{-4em} \text{Ca}^{ii+}(aq) + two\text{Cl}^{-}(aq) + 2\text{Ag}^{+}(aq) + \rule[0.5ex]{4.5em}{0.1ex}\hspace{-4.5em} 2\text{NO}_3^{-}(aq) \longrightarrow \rule[0.5ex]{4em}{0.1ex}\hspace{-4em} \text{Ca}^{2+}(aq) + \dominion[0.5ex]{4.5em}{0.1ex}\hspace{-iv.5em} two{\text{NO}_3}^{-}(aq) + two\text{AgCl}(s)[/latex]

[latex]2\text{Cl}^{-}(aq) + 2\text{Ag}^{+}(aq) \longrightarrow 2\text{AgCl}(s)[/latex]

Following the convention of using the smallest possible integers as coefficients, this equation is and so written:

[latex]\text{Cl}^{-}(aq) + \text{Ag}^{+}(aq) \longrightarrow \text{AgCl}(s)[/latex]

This net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These molecular and consummate ionic equations provide additional information, namely, the ionic compounds used as sources of Cl⁻ and Ag⁺.

Case 2

Molecular and Ionic Equations
When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Write counterbalanced molecular, complete ionic, and internet ionic equations for this process.

Solution
Begin by identifying formulas for the reactants and products and arranging them properly in chemical equation class:

[latex]\text{CO}_2(aq) + \text{NaOH}(aq) \longrightarrow \text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{O}(l) \;(\text{unbalanced})[/latex]

Balance is achieved easily in this instance by changing the coefficient for NaOH to ii, resulting in the molecular equation for this reaction:

[latex]\text{CO}_2(aq) + 2\text{NaOH}(aq) \longrightarrow \text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{O}(l)[/latex]

The two dissolved ionic compounds, NaOH and Na₂CO_iii, tin can be represented as dissociated ions to yield the complete ionic equation:

[latex]\text{CO}_2(aq) + 2\text{Na}^{+}(aq) + 2\text{OH}^{-}(aq) \longrightarrow two\text{Na}^{+}(aq) + {\text{CO}_3}^{2-}(aq) + \text{H}_2 \text{O}(l)[/latex]

Finally, identify the spectator ion(due south), in this case Na⁺(aq), and remove it from each side of the equation to generate the net ionic equation:

[latex]\text{CO}_2(aq) + \rule[0.5ex]{4.25em}{0.1ex}\hspace{-4.25em} 2\text{Na}^{+}(aq) + 2\text{OH}^{-}(aq) \longrightarrow \dominion[0.5ex]{4.25em}{0.1ex}\hspace{-4.25em} 2\text{Na}^{+}(aq) + {\text{CO}_3}^{2-}(aq) + \text{H}_2 \text{O}(l)[/latex][latex]\text{CO}_2(aq) + two\text{OH}^{-}(aq) \longrightarrow {\text{CO}_3}^{2-}(aq) + \text{H}_2 \text{O}(l)[/latex]

Check Your Learning
Diatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen, via the electrolysis of brine, according to the following unbalanced equation:

[latex]\text{NaCl}(aq) + \text{H}_2 \text{O}(l) \;\;\xrightarrow{\text{electricity}}\;\; \text{NaOH}(aq) + \text{H}_2(g) + \text{Cl}_2(one thousand)[/latex]

Write counterbalanced molecular, complete ionic, and net ionic equations for this process.

Answer:

[latex]2\text{NaCl}(aq) + 2\text{H}_2 \text{O} \longrightarrow 2 \text{NaOH}(aq) + \text{H}_2(g) + \text{Cl}_2(yard) (\text{molecular})[/latex]

[latex]2\text{Na}^{+}(aq) + 2\text{Cl}^{-}(aq) + 2\text{H}_2 \text{O} \longrightarrow 2\text{Na}^{+}(aq) + 2\text{OH}^{-}(aq) + \text{H}_2(yard) + \text{Cl}_2(chiliad) (\text{complete ionic})[/latex]

[latex]two\text{Cl}^{-}(aq) + 2\text{H}_2 \text{O} \longrightarrow two\text{OH}^{-}(aq) + 2\text{H}_2(k) + \text{Cl}_2(m) (\text{cyberspace ionic})[/latex]

Key Concepts and Summary

Chemical equations are symbolic representations of chemic and concrete changes. Formulas for the substances undergoing the change (reactants) and substances generated by the alter (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products. Chemical reactions in aqueous solution that involve ionic reactants or products may be represented more realistically by complete ionic equations and, more succinctly, past net ionic equations.

Chemistry End of Chapter Exercises

1. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?
2. Consider molecular, complete ionic, and net ionic equations.

   (a) What is the deviation between these types of equations?

   (b) In what circumstance would the complete and net ionic equations for a reaction be identical?

3. Remainder the post-obit equations:

   (a) [latex]\text{PCl}_5(s) + \text{H}_2 \text{O}(l) \longrightarrow \text{POCl}_3(l) + \text{HCl}(aq)[/latex]

   (b) [latex]\text{Cu}(s) + \text{HNO}_3(aq) \longrightarrow \text{Cu(NO}_3)_2(aq) + \text{H}_2 \text{O}(l) + \text{NO}(g)[/latex]

   (c) [latex]\text{H}_2(g) + \text{I}_2(southward) \longrightarrow \text{HI}(s)[/latex]

   (d) [latex]\text{Fe}(s) + \text{O}_2(g) \longrightarrow \text{Fe}_2 \text{O}_3(southward)[/latex]

   (e) [latex]\text{Na}(south) + \text{H}_2 \text{O}(50) \longrightarrow \text{NaOH}(aq) + \text{H}_2(g)[/latex]

   (f) [latex]\text{(NH}_4)_2 \text{Cr}_2\text{O}_7(southward) \longrightarrow \text{Cr}_2\text{O}_3(s) + \text{North}_2(g) + \text{H}_2 \text{O}(yard)[/latex]

   (one thousand) [latex]\text{P}_4(due south) + \text{Cl}_2(g) \longrightarrow \text{PCl}_3(l)[/latex]

   (h) [latex]\text{PtCl}_4(s) \longrightarrow \text{Pt}(due south) + \text{Cl}_2(1000)[/latex]

4. Balance the following equations:

   (a) [latex]\text{Ag}(southward) + \text{H}_2 \text{S}(k) + \text{O}_2(chiliad) \longrightarrow \text{Ag}_2 \text{Due south}(s) + \text{H}_2 \text{O}(fifty)[/latex]

   (b) [latex]\text{P}_4(s) + \text{O}_2(g) \longrightarrow \text{P}_4 \text{O}_{10}(southward)[/latex]

   (c) [latex]\text{Pb}(s) + \text{H}_2 \text{O}(l) + \text{O}_2(g) \longrightarrow \text{Lead(OH)}_2(s)[/latex]

   (d) [latex]\text{Fe}(s) + \text{H}_2 \text{O}(l) \longrightarrow \text{Fe}_3 \text{O}_4(south) + \text{H}_2(chiliad)[/latex]

   (east) [latex]\text{Sc}_2 \text{O}_3(s) + \text{Then}_3(50) \longrightarrow \text{Sc}_2 \text{(Then}_4)_3(s)[/latex]

   (f) [latex]\text{Ca}_3 \text{(PO}_4)_2(aq) + \text{H}_3 \text{PO}_4(aq) \longrightarrow \text{Ca(H}_2 \text{PO}_4)_2(aq)[/latex]

   (g) [latex]\text{Al}(s) + \text{H}_2 \text{SO}_4(aq) \longrightarrow \text{Al}_2 \text{(And then}_4)_3(south) + \text{H}_2(g)[/latex]

   (h) [latex]\text{TiCl}_4(s) + \text{H}_2 \text{O}(g) \longrightarrow \text{TiO}_2(s) + \text{HCl}(g)[/latex]

5. Write a balanced molecular equation describing each of the post-obit chemic reactions.

   (a) Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.

   (b) Gaseous butane, C₄H₁₀, reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.

   (c) Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.

   (d) Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.

6. Write a balanced equation describing each of the following chemical reactions.

   (a) Solid potassium chlorate, KClO₃, decomposes to form solid potassium chloride and diatomic oxygen gas.

   (b) Solid aluminum metal reacts with solid diatomic iodine to form solid Al_twoI₆.

   (c) When solid sodium chloride is added to aqueous sulfuric acid, hydrogen chloride gas and aqueous sodium sulfate are produced.

   (d) Aqueous solutions of phosphoric acid and potassium hydroxide react to produce aqueous potassium dihydrogen phosphate and liquid water.

7. Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and fe with oxygen.

   (a) Write the formulas of barium nitrate and potassium chlorate.

   (b) The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.

   (c) The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.

   (d) Write separate equations for the reactions of the solid metals magnesium, aluminum, and fe with diatomic oxygen gas to yield the respective metal oxides. (Assume the fe oxide contains Atomic number 26^three+ ions.)

8. Fill in the blank with a single chemical formula for a covalent compound that will balance the equation:
9. Aqueous hydrogen fluoride (hydrofluoric acrid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride volition also react with sand (silicon dioxide).

   (a) Write an equation for the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid h2o.

   (b) The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride tin also be prepared by the reaction of aqueous solutions of calcium chloride and sodium fluoride, yielding aqueous sodium chloride equally the other production. Write consummate and net ionic equations for this reaction.

10. A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each footstep of the process.

    (a) The kickoff pace is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide.

    (b) The second footstep is the germination of solid calcium hydroxide as the only product from the reaction of the solid calcium oxide with liquid water.

    (c) Solid calcium hydroxide is then added to the seawater, reacting with dissolved magnesium chloride to yield solid magnesium hydroxide and aqueous calcium chloride.

    (d) The solid magnesium hydroxide is added to a hydrochloric acid solution, producing dissolved magnesium chloride and liquid water.

    (due east) Finally, the magnesium chloride is melted and electrolyzed to yield liquid magnesium metal and diatomic chlorine gas.

11. From the balanced molecular equations, write the complete ionic and net ionic equations for the post-obit:

    (a) [latex]\text{Grand}_2 \text{C}_2 \text{O}_4(aq) + \text{Ba(OH)}_2(aq) \longrightarrow two\text{KOH}(aq) + \text{BaC}_2 \text{O}_2(s)[/latex]

    (b) [latex]{\text{Lead(NO}_3)}_2(aq) + \text{H}_2 \text{And then}_4(aq) \longrightarrow \text{PbSO}_4(s) + two\text{HNO}_3(aq)[/latex]

    (c) [latex]\text{CaCO}_3(s) + \text{H}_2 \text{And then}_4(aq) \longrightarrow \text{CaSO}_4(s) + \text{CO}_2(chiliad) + \text{H}_2\text{O}(50)[/latex]

Glossary

balanced equation

chemic equation with equal numbers of atoms for each chemical element in the reactant and product

chemical equation

symbolic representation of a chemical reaction

coefficient

number placed in forepart of symbols or formulas in a chemic equation to indicate their relative corporeality

complete ionic equation

chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions

molecular equation

chemic equation in which all reactants and products are represented equally neutral substances

net ionic equation

chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical alter are represented (excludes spectator ions)

product

substance formed by a chemical or concrete change; shown on the right side of the pointer in a chemical equation

reactant

substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation

spectator ion

ion that does not undergo a chemical or concrete change during a reaction, but its presence is required to maintain accuse neutrality

Solutions

Answers to Chemistry Finish of Chapter Exercises

1. An equation is balanced when the aforementioned number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.

3.

(a) [latex]\text{PCl}_5(southward) + \text{H}_2 \text{O}(l) \longrightarrow \text{POCl}_3(l) + 2\text{HCl}(aq)[/latex];

(b) [latex]three\text{Cu}(southward) + 8\text{HNO}_3(aq) \longrightarrow iii\text{Cu(NO}_3)_2(aq) + 4\text{H}_2 \text{O}(l) + two\text{NO}(thou)[/latex];

(c) [latex]\text{H}_2(g) + \text{I}_2(south) \longrightarrow two\text{HI}(s)[/latex];

(d) [latex]4\text{Fe}(s) + 3\text{O}_2(grand) \longrightarrow two\text{Atomic number 26}_2 \text{O}_3(southward)[/latex];

(e) [latex]two\text{Na}(s) + two\text{H}_2 \text{O}(50) \longrightarrow ii\text{NaOH}(aq) + \text{H}_2(g)[/latex];

(f) [latex]\text{(NH}_4)_2 \text{Cr}_2\text{2O}_7(s) \longrightarrow \text{Cr}_2\text{O}_3(s) + \text{Due north}_2(thousand) + four\text{H}_2 \text{O}(l)[/latex];

(g) [latex]\text{P}_4(s) + 6\text{Cl}_2(g) \longrightarrow four\text{PCl}_3(l)[/latex];

(h) [latex]\text{PtCl}_4(s) \longrightarrow \text{Pt}(s) + ii\text{Cl}_2(g)[/latex];

five.
(a) [latex]\text{CaCO}_3(s) \longrightarrow \text{CaO}(due south) + \text{CO}_2(m)[/latex];
(b) [latex]2\text{C}_4 \text{H}_{10}(chiliad) + 13 \text{O}_2(g) \longrightarrow 8\text{CO}_2(chiliad) + 10\text{H}_2\text{O}(g)[/latex];
(c) [latex]\text{MgCl}_{two}(aq) + 2 \text{NaOH}(aq) \longrightarrow \text{Mg(OH)}_2(s) + 2 \text{NaCl}(aq)[/latex];
(d) [latex]two\text{H}_2 \text{O}(g) + 2 \text{Na}(s) \longrightarrow 2\text{NaOH}(s) + \text{H}_2(g)[/latex];

7.
(a) [latex]\text{Ba(NO}_3)_2[/latex] , [latex]\text{KClO}_3[/latex];
(b) [latex]2 \text{KClO}_3(south) \longrightarrow 2 \text{KCl}(s) + 3\text{O}_2(m)[/latex] ;
(c) [latex]2 \text{Ba(NO}_3)_2(due south) \longrightarrow ii\text{BaO}(southward) + 2\text{Due north}_2(g) + 5\text{O}_2(g)[/latex] ;
(d) [latex]2 \text{Mg}(s) + \text{O}_2(g) \longrightarrow two \text{MgO}(s)[/latex]; [latex]4\text{Al}(s) + 3\text{O}_2(g) \longrightarrow 2\text{Al}_2 \text{O}_3(1000)[/latex];[latex]4\text{Fe}(south) + 3\text{O}_2(g) \longrightarrow two\text{Fe}_2 \text{O}_3(southward)[/latex];

9.
(a) [latex]4\text{HF}(aq) + \text{SiO}_2(s) \longrightarrow \text{SiF}_4(1000) + 2\text{H}_2 \text{O}(l)[/latex];
(b) complete ionic equation: [latex]2\text{Na}^{+}(aq) + 2\text{F}^{-}(aq) + \text{Ca}^{2+}(aq) + 2\text{Cl}^{-}(aq) \longrightarrow \text{CaF}_2(s) + 2\text{Na}^{+}(aq) + 2\text{Cl}^{-}(aq)[/latex]
net ionic equation: [latex]2\text{F}^{-}(aq) + \text{Ca}^{two+}(aq) \longrightarrow \text{CaF}_2(due south)[/latex]

xi.
(a)
[latex]2\text{K}^{+}(aq) + {\text{C}_2 \text{O}_4}^{ii-}(aq) + \text{Ba}^{two+}(aq) + 2\text{OH}^{-}(aq) \longrightarrow 2\text{Yard}^{+}(aq) + two\text{OH}^{-}(aq) + \text{BaC}_2 \text{O}_4(s) \;\text{(complete)}[/latex]
[latex]\text{Ba}^{two+}(aq) + {\text{C}_2 {\text{O}_4}}^{2-}(aq) \longrightarrow \text{BaC}_2 \text{O}_4(south) \;(\text{net})[/latex]

(b)
[latex]\text{Lead}^{2+}(aq) + 2{\text{NO}_3}^{-}(aq) + 2\text{H}^{+}(aq) + {\text{And then}_4}^{two-}(aq) \longrightarrow \text{PbSO}_4(south) + 2\text{H}^{+}(aq) + ii{\text{NO}_3}^{-}(aq) \;\text{(complete)}[/latex]

[latex]\text{Pb}^{2+}(aq) + {\text{SO}_4}^{2-}(aq) \longrightarrow \text{PbSO}_4(s) \;\text{(net)}[/latex]

(c)
[latex]\text{CaCO}_3(s) + 2\text{H}^{+}(aq) + {\text{Then}_4}^{2-}(aq) \longrightarrow \text{CaSO}_4(s) + \text{CO}_2(g) + \text{H}_2 \text{O}(50) \;(\text{consummate})[/latex]

[latex]\text{CaCO}_3(southward) + 2\text{H}^{+}(aq) + {\text{SO}_4}^{2-}(aq) \longrightarrow \text{CaSO}_4(s) + \text{CO}_2(g) + \text{H}_2 \text{O}(50) \;(\text{net})[/latex]

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Source: https://opentextbc.ca/chemistry/chapter/4-1-writing-and-balancing-chemical-equations/

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