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Chapter 11: Problem 4

Challenge Balance the following equations, and determine the possible mole ratios. a. \(\mathrm{ZnO}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) b. butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)+\) oxygen \(\rightarrow\) carbon dioxide \(+\) water

Short Answer

Expert verified
The balanced chemical equations and their mole ratios are: (a) \(\mathrm{ZnO}(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) Mole ratios: ZnO:HCl:ZnCl2:H2O = 1:2:1:1 (b) \(2\mathrm{C}_{4} \mathrm{H}_{10}(g) + 13\mathrm{O}_{2}(g) \rightarrow 8\mathrm{CO}_{2}(g) + 10\mathrm{H}_{2} \mathrm{O}(g)\) Mole ratios: C4H10:O2:CO2:H2O = 1:6.5:4:5

Step by step solution

01

Balance equation (a)

First, let's balance the first equation. The unbalanced equation is given by: \(\mathrm{ZnO}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) Now, let's balance the atoms in the equation: 1. Balance the Zinc (Zn) atoms: They are already balanced, as there is 1 Zn atom on both sides. 2. Balance the Oxygen (O) atoms: They are also balanced, with 1 O atom on both sides. 3. Balance the Hydrogen (H) and Chlorine (Cl) atoms: There are 2 H atoms and 2 Cl atoms on the right side of the equation, so we must multiply HCl by 2. The balanced equation is: \(\mathrm{ZnO}(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
02

Determine mole ratios for equation (a)

Based on the balanced chemical equation: \(\mathrm{ZnO}(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) The mole ratios are as follows: 1. ZnO: \(\frac{1}{1}\) = 1 2. HCl: \(\frac{2}{1}\) = 2 3. ZnCl2: \(\frac{1}{1}\) = 1 4. H2O: \(\frac{1}{1}\) = 1 We write all of these as mole ratios: ZnO:HCl:ZnCl2:H2O = 1:2:1:1
03

Balance equation (b)

Now, let's balance the second equation, involving butane and oxygen. The unbalanced equation is given by: \(\mathrm{C}_{4} \mathrm{H}_{10}(g) + \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(g)\) Now, let's balance the atoms in the equation: 1. Balance the Carbon (C) atoms: We have 4 C atoms on the left side and 1 C atom on the right side, so we must multiply CO2 by 4. 2. Balance the Hydrogen (H) atoms: We have 10 H atoms on the left side and 2 H atoms on the right side, so we must multiply H2O by 5. 3. Balance the Oxygen (O) atoms: Now we have 8 O atoms from CO2 and 5 O atoms from H2O, which totals 13 O atoms on the right side. To balance this, we must multiply O2 by 6.5 or \(\frac{13}{2}\) on the left side. However, we should avoid having fractions in our balanced equation, so we can multiply the entire equation by 2 to have whole numbers. The balanced equation is: \(2\mathrm{C}_{4} \mathrm{H}_{10}(g) + 13\mathrm{O}_{2}(g) \rightarrow 8\mathrm{CO}_{2}(g) + 10\mathrm{H}_{2} \mathrm{O}(g)\)
04

Determine mole ratios for equation (b)

Based on the balanced chemical equation: \(2\mathrm{C}_{4} \mathrm{H}_{10}(g) + 13\mathrm{O}_{2}(g) \rightarrow 8\mathrm{CO}_{2}(g) + 10\mathrm{H}_{2} \mathrm{O}(g)\) The mole ratios are as follows: 1. C4H10: \(\frac{2}{2}\) = 1 2. O2: \(\frac{13}{2}\) = 6.5 3. CO2: \(\frac{8}{2}\) = 4 4. H2O: \(\frac{10}{2}\) = 5 We write all of these as mole ratios: C4H10:O2:CO2:H2O = 1:6.5:4:5

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mole Ratios
Understanding mole ratios is crucial when studying chemistry, as it involves the proportions of reactants and products in a chemical reaction. Mole ratios indicate the relative amounts of each substance involved in the reaction. For example, in the balanced equation \(\text{ZnO}(\text{s}) + 2\text{HCl}(\text{aq}) \to \text{ZnCl}_2(\text{aq}) + \text{H}_2 \text{O}(\text{l})\), the mole ratio can be interpreted as follows:
  • For every 1 mole of \text{ZnO} used, 2 moles of \text{HCl} are needed.
  • 1 mole of \text{ZnO} produces 1 mole of \text{ZnCl}_2 and 1 mole of \text{H}_2O.
The simplicity of these ratios allows us to predict the amounts of chemicals we need for a reaction or the amount we can expect to produce in a product. Similarly, in the balanced reaction involving butane, by understanding the mole ratios, chemists can calculate the exact amount of oxygen needed to fully combust a certain volume of butane, making mole ratios an indispensable tool in practical chemistry.
Stoichiometry
Stoichiometry might sound like a complex term, but it's essentially the arithmetic behind chemistry. It deals with the quantification of the substances before and after the chemical reactions take place, using the mole ratios previously discussed. If you're following a cooking recipe, you need to know how much of each ingredient to add to get the desired number of servings; stoichiometry is the 'recipe' for chemical reactions.

For instance, if a chemist wants to produce water, \( \text{H}_2\text{O} \), knowing that the mole ratio of \( \text{ZnO}:\text{HCl}:\text{ZnCl}_2:\text{H}_2\text{O} \) is 1:2:1:1 ensures that no \text{ZnO} or \text{HCl} is left unreacted, which is highly efficient. Stoichiometry helps us to conserve resources and reduce waste, both economically and environmentally beneficial.
Chemical Reactions
Chemical reactions are the transformations that alter the composition of substances. In a chemical reaction, the reactants convert into products, which have different properties from the reactants. For example, in the reaction where \( \text{ZnO} \) and \( \text{HCl} \) react to form \( \text{ZnCl}_2 \) and water, \( \text{ZnO} \) (a metal oxide) and \( \text{HCl} \) (a strong acid) result in an aqueous salt solution and water鈥攁 totally different substance from the initial reactants.

To visualize chemical reactions, you must be able to read and write chemical equations, which represent these transformations symbolically. The balancing of chemical equations, as demonstrated in the original exercise, is a fundamental skill in chemistry. It's based on the law of mass conservation, which states that mass can neither be created nor destroyed in a reaction, meaning that all atoms present in the reactants must be accounted for in the products.

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Most popular questions from this chapter

Challenge When copper wire is placed into a silver nitrate solution \(\left(\mathrm{AgNO}_{3}\right)\), silver crystals and copper(II) nitrate \(\left(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\right)\) solution form. a. Write the balanced chemical equation for the reaction. b. If a 20.0-g sample of copper is used, determine the theoretical yield of silver. c. If 60.0 g of silver is recovered from the reaction, determine the percent yield of the reaction.

Car Battery Car batteries use lead, lead(IV) oxide, and a sulfuric acid solution to produce an electric current. The products of the reaction are lead(II) sulfate in solution and water. a. Write the balanced equation for the reaction. b. Determine the mass of lead(II) sulfate produced when 25.0 g of lead reacts with an excess of lead(IV) oxide and sulfuric acid.

Solid silicon dioxide, often called silica, reacts with hydrofluoric acid (HF) solution to produce the gas silicon tetrafluoride and water. a. Write the balanced chemical equation for the reaction. b. List three mole ratios, and explain how you would use them in stoichiometric calculations.

Lead(II) oxide is obtained by roasting galena, lead(II) sulfide, in air. The unbalanced equation is: $$\mathrm{PbS}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{PbO}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g})$$ a. Balance the equation, and determine the theoretical yield of PbO if 200.0 g of PbS is heated. b. What is the percent yield if 170.0 g of PbO is obtained?

When hydrochloric acid solution reacts with lead(II) nitrate solution, lead(II) chloride precipitates and a solution of nitric acid is produced. a. Write the balanced chemical equation for the reaction. b. Interpret the equation in terms of molecules and formula units, moles, and mass.
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