1. when a potassium chlorate sample was heated in the presence of a catalyst, $25.7\text{ ml}$ of $\text{o}_2(\text{g})$ was collected over water at $14^\circ\text{c}$ and $787\text{ mm hg}$. what was the mass of $\text{kclo}_3$ in the sample?

\2\text{ kclo}_3(\text{s}) \
ightarrow 2\text{ kcl}(\text{s}) + 3\text{ o}_2(\text{g})\

Question

1. when a potassium chlorate sample was heated in the presence of a catalyst, $25.7\text{ ml}$ of $\text{o}_2(\text{g})$ was collected over water at $14^\circ\text{c}$ and $787\text{ mm hg}$. what was the mass of $\text{kclo}_3$ in the sample?

\2\text{ kclo}_3(\text{s}) \
ightarrow 2\text{ kcl}(\text{s}) + 3\text{ o}_2(\text{g})\

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Pressure Unit Conversion", "Ideal Gas Law Calculations", "Gas Stoichiometry", "Molar Mass Calculation" ], "new_concepts": [ "Gas Collection Over Water", "Vapor Pressure of Water" ], "current_concepts": [ "Pressure Unit Conversion", "Ideal Gas Law Calculations", "Gas Stoichiometry", "Molar Mass Calculation", "Gas Collection Over Water", "Vapor Pressure of Water" ] } ## Calculate partial pressure of oxygen Using the *Pressure Unit Conversion* knowledge point $$ \begin{aligned} &T = 14^\circ\text{C} = 287.15\text{ K}\ &P_{\text{H}_2\text{O}} = 12.0\text{ mm Hg}\quad (\text{at } 14^\circ\text{C})\ &P_{\text{O}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}} = 787\text{ mm Hg} - 12.0\text{ mm Hg} = 775\text{ mm Hg}\ &P_{\text{O}_2} = \frac{775\text{ mm Hg}}{760\text{ mm Hg/atm}} \approx 1.0197\text{ atm} \end{aligned} $$ ## Calculate moles of oxygen gas Using the *Ideal Gas Law Calculations* knowledge point $$ \begin{aligned} &V = 25.7\text{ mL} = 0.0257\text{ L}\ &n_{\text{O}_2} = \frac{P_{\text{O}_2} V}{R T} = \frac{(1.0197\text{ atm})(0.0257\text{ L})}{(0.08206\text{ L}\cdot\text{atm}/\text{mol}\cdot\text{K})(287.15\text{ K})} \approx 1.112 \times 10^{-3}\text{ mol} \end{aligned} $$ ## Calculate moles of potassium chlorate Using the *Gas Stoichiometry* knowledge point $$ \begin{aligned} &2\text{ KClO}_3(\text{s}) ightarrow 2\text{ KCl}(\text{s}) + 3\text{ O}_2(\text{g})\ &n_{\text{KClO}_3} = n_{\text{O}_2} \times \frac{2\text{ mol KClO}_3}{3\text{ mol O}_2} = 1.112 \times 10^{-3}\text{ mol} \times \frac{2}{3} \approx 7.413 \times 10^{-4}\text{ mol} \end{aligned} $$ ## Calculate mass of potassium chlorate Using the *Molar Mass Calculation* knowledge point $$ \begin{aligned} &M_{\text{KClO}_3} = 39.10 + 35.45 + 3(16.00) = 122.55\text{ g/mol}\ &m_{\text{KClO}_3} = n_{\text{KClO}_3} \times M_{\text{KClO}_3} = 7.413 \times 10^{-4}\text{ mol} \times 122.55\text{ g/mol} \approx 0.0908\text{ g} \end{aligned} $$ $0.0908\text{ g}$ { "subject": "Natural Science", "question_type": "Free Response", "knowledge_point": [ "Natural Science", "Chemistry", "Gas Stoichiometry" ] }

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Explanation:

Calculate partial pressure of oxygen

Calculate moles of oxygen gas

Calculate moles of potassium chlorate

Calculate mass of potassium chlorate

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