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Chapter 8: Problem 32

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts. $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) $$ (a) \(2.72 \mathrm{~g} \mathrm{KClO}_{3}\) (b) \(0.361 \mathrm{~g} \mathrm{KClO}_{3}\) (c) \(83.6 \mathrm{~kg} \mathrm{KClO}_{3}\) (d) \(22.4 \mathrm{mg} \mathrm{KClO}_{3}\)

Short Answer

Expert verified
(a) 1.07 g, (b) 0.141 g, (c) 65.28 kg, (d) 8.77 mg of oxygen gas.

Step by step solution

01

Find the Molar Mass of KClO3

Calculate the molar mass of KClO3 by adding the atomic masses of potassium (K), chlorine (Cl), and oxygen (O). The atomic masses for K, Cl, and O are approximately 39, 35.5, and 16 grams/mol respectively. Thus, the molar mass of KClO3 is given as: (39) + (35.5) + 3(16) = 122.5 g/mol.
02

Convert Grams of KClO3 to Moles

To convert the mass of KClO3 to moles, use the molar mass from Step 1:(a) \(\frac{2.72\text{ g}}{122.5\text{ g/mol}}\) moles(b) \(\frac{0.361\text{ g}}{122.5\text{ g/mol}}\) moles(c) First convert kg to g: \(83.6\text{ kg} = 83600\text{ g}\), then \(\frac{83600\text{ g}}{122.5\text{ g/mol}}\) moles(d) First convert mg to g: \(22.4\text{ mg} = 0.0224\text{ g}\), then \(\frac{0.0224\text{ g}}{122.5\text{ g/mol}}\) moles
03

Calculate Moles of O2 Produced

Using the stoichiometry of the reaction, calculate the moles of O2 produced. For every 2 moles of KClO3, 3 moles of O2 are produced. Thus:(a) \(\frac{3}{2} \times \text{moles of KClO3}\)(b) \(\frac{3}{2} \times \text{moles of KClO3}\)(c) \(\frac{3}{2} \times \text{moles of KClO3}\)(d) \(\frac{3}{2} \times \text{moles of KClO3}\)
04

Convert Moles of O2 to Grams

Calculate the grams of O2 using the molar mass of O2 (32 g/mol). Multiply the moles of O2 by the molar mass: (a) \(\text{moles of O2} \times 32\text{ g/mol}\)(b) \(\text{moles of O2} \times 32\text{ g/mol}\)(c) \(\text{moles of O2} \times 32\text{ g/mol}\)(d) \(\text{moles of O2} \times 32\text{ g/mol}\)
05

Perform the Calculations for Each Case

(a) \(\frac{2.72}{122.5} \times \(\frac{3}{2}\) \times 32 = 1.07\text{ g}\)(b) \(\frac{0.361}{122.5} \times \(\frac{3}{2}\) \times 32 = 0.141\text{ g}\)(c) \(\frac{83600}{122.5} \times \(\frac{3}{2}\) \times 32 = 65280\text{ g}\) or 65.28\text{ kg}(d) \(\frac{0.0224}{122.5} \times \(\frac{3}{2}\) \times 32 = 0.00877\text{ g}\) or 8.77\text{ mg}
06

State Final Answers

Convert the results from grams to the necessary unit if required (kg or mg):(a) 1.07 g O2(b) 0.141 g O2(c) 65.28 kg O2(d) 8.77 mg O2

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Key Concepts

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

Molar Mass Calculation
Understanding the concept of molar mass is crucial in stoichiometry problems. It is the weight of one mole of a substance, usually expressed in grams per mole (g/mol). To calculate the molar mass of a compound like potassium chlorate (KClO3), we sum up the atomic masses of each element, weighed by the number of times the element appears in the molecule.

For instance, the provided solution began with finding the molar mass of KClO3, using the atomic masses for the contributing elements: potassium (K), chlorine (Cl), and oxygen (O). By adding these values together, we get the molar mass of KClO3 as 122.5 g/mol. This step is foundational as it sets the stage for subsequent conversions between moles and grams.
Mole-to-Gram Conversion
Once we have the molar mass, we can convert between moles and grams --- a common necessity in chemistry problems. For example, to find out how many grams of a substance are present, you multiply the number of moles by the molar mass of the substance.

In the solution provided, this step involves calculating the amount of oxygen gas (O2) produced from a given mass of KClO3. After the molar masses are established, we use ratio and proportion, based on the coefficients from the balanced chemical equation to obtain the number of moles of reactants and products. This enables us to convert the mass of KClO3 to moles, a necessary step before we can use the reaction stoichiometry to find the mass of O2 produced.
Chemical Reaction Stoichiometry
Chemical reaction stoichiometry involves the quantitative relationship between the amounts of reactants and products in a chemical reaction. It's all about the mole ratio determined from the balanced chemical equation.

In the context of our example, the balanced equation shows that 2 moles of KClO3 produce 3 moles of O2. By knowing this, we can calculate the moles of O2 produced from the moles of KClO3 we calculated earlier. We achieve this by multiplying the moles of KClO3 by the stoichiometric coefficient ratio \(\frac{3}{2}\). This part of the solution is important as stoichiometry dictates the proportional relationships in terms of moles.
Mass-to-Mole Conversion
The converse of mole-to-gram conversion is just as integral. This process requires dividing the mass of a substance by its molar mass in order to determine the number of moles.

For example, if you have a substance of a known mass, you can determine the number of moles by dividing that mass by the molar mass of the substance. In the provided step-by-step solution, we see this method used to convert the given masses of KClO3 into moles, which we can then use to find the moles of O2. This conversion is vital as it links the mass of a reactant or product with the mole concept, allowing us to use stoichiometry in chemical calculations.

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Most popular questions from this chapter

Explain the relationship between the sign of \(\Delta H_{\mathrm{rxn}}\) and whether a reaction is exothermic or endothermic.

For the reaction shown, calculate how many moles of \(\mathrm{NH}_{3}\) form when each amount of reactant completely reacts. $$ 3 \mathrm{~N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4 \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g) $$ (a) \(5.3 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (b) \(2.28 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (c) \(5.8 \times 10^{-2} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (d) \(9.76 \times 10^{7} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\)

Consider the generic chemical reaction: $$ 2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow 3 \mathrm{C} $$ How many moles of B are required to completely react with: (a) \(6 \mathrm{~mol}\) of \(\mathrm{A}\) (b) \(2 \mathrm{~mol}\) of \(\mathrm{A}\) (c) \(7 \mathrm{~mol}\) of \(\mathrm{A}\) (d) 11 mol of \(\mathrm{A}\)

Consider the combustion of propane: $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Balance the reaction. (b) Divide all coefficients by the coefficient on propane, so that you have the reaction for the combustion of 1 mol of propane. (c) \(\Delta H_{\mathrm{rxn}}\) for the combustion of one mole of propane is \(-2219 \mathrm{~kJ}\). What mass of propane would you need to burn to generate \(5.0 \mathrm{MJ}\) of heat? (d) If propane costs about \(\$ 0.67 / \mathrm{L}\) and has a density of \(2.01 \mathrm{~g} / \mathrm{cm}^{3}\), how much would it cost to generate \(5.0 \mathrm{MJ}\) of heat by burning propane?

Consider the reaction between sulfur trioxide and water: $$ \mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ A chemist allows \(61.5 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(11.2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) to react. When the reaction is finished, the chemist collects \(54.9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Determine the limiting reactant, theoretical yield, and percent yield for the reaction.
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