Chapter 9: Problem 27

Use the equation provided to answer the questions that follow. The density of oxygen gas is 1.428 $\mathrm{g} / \mathrm{L}$ . $$2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$$ \begin{equation}\begin{array}{l}{\text { a. What volume of oxygen can be made }} \\ {\text { from } 5.00 \times 10^{-2} \text { mol of } \mathrm{KClO}_{3} ?} \\ {\text { b. How many grams } \mathrm{KClO}_{3} \text { must react to }} \\\ {\text { form } 42.0 \mathrm{mL} \mathrm{O}_{2} \text { ? }} \\ {\text { c. How many milliliters of } \mathrm{O}_{2} \text { will form at }} \\ {\text { STP from } 55.2 \mathrm{g} \mathrm{KClO}_{3} ?}\end{array}\end{equation}

Short Answer

Expert verified

a. The volume of oxygen that can be produced by 5.00 x 10^-2 mol of $KClO_{3}$ is 1.68 L. b. To form 42.0 mL of $O_{2}$, 0.040 g $KClO_{3}$ must react. c. 55.2 g of $KClO_{3}$ will form 15.1 L of $O_{2}$ at STP.

Step by step solution

Determine the molar ratio and the volume of oxygen formed

From the balanced chemical equation, it is evident that for every 2 moles of $KClO_{3}$, 3 moles of $O_{2}$ are produced. Thus, if 0.05 moles of $KClO_{3}$ reacts, the moles of $O_{2}$ formed, by cross multiplication, will be $0.05 \times 3 / 2=0.075$ moles. At STP, 1 mole of any gas occupies 22.4 L. Therefore, volume of $O_{2}$ formed is $0.075 \times 22.4 = 1.68 L$

Determine the number of grams of $KClO_{3}$ that must react

Given that 1.428 g of $O_{2}$ occupies 1 L, the mass of 42.0 mL (or $0.042 L$ of $O_{2}$ is $1.428 \times 0.042 = 0.060 g$. From the stoichiometry of the reaction, 3 moles of $O_{2}$ is produced from 2 moles of $KClO_{3}$. Therefore, the mass of $KClO_{3}$ that should react to form 0.060 g $O_{2}$ is $0.060 \times 2 / 3 = 0.040 g$.

Determine volume of $O_{2}$ formed from specific mass $KClO_{3}$

First, convert mass of $KClO_{3}$ to moles using its molar mass $K=39.1 g/mol$, $Cl=35.5 g/mol$, $O=16.0 g/mol$. So, molar mass of $KClO_{3}=(39.1+35.5+(3 \times 16.0))g/mol=122.6 g/mol$. Then, moles of $KClO_{3}=55.2 g/122.6 g per mol= 0.45 mol$. From the stoichiometry, 3 moles $O_{2}$ is produced from 2 moles of $KClO_{3}$. Therefore, moles of $O_{2}$ produced is $0.45 \times 3 / 2 = 0.675 mol$. Finally, volume of $O_{2}$ produced at STP is $0.675 mol \times 22.4 L/mol = 15.1 L$

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Short Answer

Step by step solution

Determine the molar ratio and the volume of oxygen formed

Determine the number of grams of \(KClO_{3}\) that must react

Determine volume of \(O_{2}\) formed from specific mass \(KClO_{3}\)

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