when powdered zinc is heated with sulfur, a violent reaction occurs, and zinc sulfide forms:

$\text{zn}(s) + \text{s}_8(s) \
ightarrow \text{zns}(s)$ unbalanced

some of the reactants also combine with oxygen in air to form zinc oxide and sulfur dioxide. when $83.2\text{ g}$ of $\text{zn}$ reacts with $52.4\text{ g}$ of $\text{s}_8$, $109.2\text{ g}$ of $\text{zns}$ forms.

part 1 of 3
what is the percent yield of $\text{zns}$? be sure your answer has the correct number of significant figures.

part 2 of 3
if all the remaining reactants combine with oxygen, how many grams of $\text{zno}$ form? be sure your answer has the correct number of significant figures.

part 3 of 3
if all the remaining reactants combine with oxygen, how many grams of $\text{so}_2$ form? be sure your answer has the correct number of significant figures.

Question

when powdered zinc is heated with sulfur, a violent reaction occurs, and zinc sulfide forms:

$\text{zn}(s) + \text{s}_8(s) \
ightarrow \text{zns}(s)$ unbalanced

some of the reactants also combine with oxygen in air to form zinc oxide and sulfur dioxide. when $83.2\text{ g}$ of $\text{zn}$ reacts with $52.4\text{ g}$ of $\text{s}_8$, $109.2\text{ g}$ of $\text{zns}$ forms.

part 1 of 3
what is the percent yield of $\text{zns}$? be sure your answer has the correct number of significant figures.

part 2 of 3
if all the remaining reactants combine with oxygen, how many grams of $\text{zno}$ form? be sure your answer has the correct number of significant figures.

part 3 of 3
if all the remaining reactants combine with oxygen, how many grams of $\text{so}_2$ form? be sure your answer has the correct number of significant figures.

Accepted Answer

{ "quality": "clear", "question_count": 3, "skills_matched": [ "step_cot", "explain_then_answer", "tutor" ], "needs_drawing": false, "concepts_used": [ "Chemical Equation Balancing", "Stoichiometry", "Percent Yield Calculations", "Limiting Reactant Stoichiometry" ], "new_concepts": [], "current_concepts": [ "Chemical Equation Balancing", "Stoichiometry", "Percent Yield Calculations", "Limiting Reactant Stoichiometry" ] } ## Balance the chemical equation The given unbalanced equation is: $$\text{Zn}(s) + \text{S}_8(s) ightarrow \text{ZnS}(s)$$ To balance the sulfur atoms, we need 8 molecules of $\text{ZnS}$ on the product side, which in turn requires 8 atoms of $\text{Zn}$ on the reactant side: $$8\text{Zn}(s) + \text{S}_8(s) ightarrow 8\text{ZnS}(s)$$ ## Determine the limiting reactant Using **Limiting Reactant Stoichiometry**, we calculate the moles of each reactant: - Molar mass of $\text{Zn} = 65.38\text{ g/mol}$ - Molar mass of $\text{S}_8 = 8 \times 32.06 = 256.48\text{ g/mol}$ Moles of reactants: $$n_{\text{Zn}} = \frac{83.2\text{ g}}{65.38\text{ g/mol}} \approx 1.2726\text{ mol}$$ $$n_{\text{S}_8} = \frac{52.4\text{ g}}{256.48\text{ g/mol}} \approx 0.2043\text{ mol}$$ From the balanced equation, 1 mole of $\text{S}_8$ reacts with 8 moles of $\text{Zn}$. The moles of $\text{Zn}$ required to react completely with $0.2043\text{ mol}$ of $\text{S}_8$ is: $$0.2043\text{ mol} \times 8 = 1.6344\text{ mol}$$ Since we only have $1.2726\text{ mol}$ of $\text{Zn}$, $\text{Zn}$ is the limiting reactant. ## Calculate the theoretical and percent yield of ZnS Using **Percent Yield Calculations** and **Stoichiometry**: Since $\text{Zn}$ is the limiting reactant, the theoretical yield of $\text{ZnS}$ is: $$n_{\text{ZnS, theoretical}} = n_{\text{Zn}} \approx 1.2726\text{ mol}$$ Molar mass of $\text{ZnS} = 65.38 + 32.06 = 97.44\text{ g/mol}$ $$m_{\text{ZnS, theoretical}} = 1.2726\text{ mol} \times 97.44\text{ g/mol} \approx 124.00\text{ g}$$ The actual yield is $109.2\text{ g}$. $$\text{Percent Yield} = \frac{109.2\text{ g}}{124.00\text{ g}} \times 100\% \approx 88.1\%$$ ## Calculate the mass of ZnO formed The actual amount of $\text{Zn}$ that reacted to form $\text{ZnS}$ is: $$n_{\text{Zn, reacted}} = \frac{109.2\text{ g}}{97.44\text{ g/mol}} \approx 1.1207\text{ mol}$$ The remaining unreacted $\text{Zn}$ is: $$n_{\text{Zn, remaining}} = 1.2726\text{ mol} - 1.1207\text{ mol} = 0.1519\text{ mol}$$ This remaining $\text{Zn}$ reacts with oxygen to form $\text{ZnO}$ according to: $$2\text{Zn} + \text{O}_2 ightarrow 2\text{ZnO}$$ Thus, $0.1519\text{ mol}$ of $\text{ZnO}$ is formed. Molar mass of $\text{ZnO} = 65.38 + 16.00 = 81.38\text{ g/mol}$ $$m_{\text{ZnO}} = 0.1519\text{ mol} \times 81.38\text{ g/mol} \approx 12.4\text{ g}$$ ## Calculate the mass of SO2 formed The amount of $\text{S}_8$ that reacted to form $\text{ZnS}$ is: $$n_{\text{S}_8\text{, reacted}} = \frac{1.1207\text{ mol ZnS}}{8} \approx 0.1401\text{ mol}$$ The remaining unreacted $\text{S}_8$ is: $$n_{\text{S}_8\text{, remaining}} = 0.2043\text{ mol} - 0.1401\text{ mol} = 0.0642\text{ mol}$$ This remaining $\text{S}_8$ reacts with oxygen to form $\text{SO}_2$ according to: $$\text{S}_8 + 8\text{O}_2 ightarrow 8\text{SO}_2$$ Thus, the moles of $\text{SO}_2$ formed is: $$n_{\text{SO}_2} = 8 \times n_{\text{S}_8\text{, remaining}} = 8 \times 0.0642\text{ mol} = 0.5136\text{ mol}$$ Molar mass of $\text{SO}_2 = 32.06 + 2 \times 16.00 = 64.06\text{ g/mol}$ $$m_{\text{SO}_2} = 0.5136\text{ mol} \times 64.06\text{ g/mol} \approx 32.9\text{ g}$$ ## Question 1 The percent yield of $\text{ZnS}$ is 88.1 $\%$. ## Question 2 The mass of $\text{ZnO}$ formed is 12.4 $\text{g}$. ## Question 3 The mass of $\text{SO}_2$ formed is 32.9 $\text{g}$. { "subject": "Natural Science", "question_type": "Multi-part", "knowledge_point": [ "Natural Science", "Chemistry", "Limiting Reactant Stoichiometry" ] }

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QUESTION IMAGE

Question

Explanation:

Balance the chemical equation

Determine the limiting reactant

Calculate the theoretical and percent yield of ZnS

Calculate the mass of ZnO formed

Calculate the mass of SO2 formed

Answer:

Question 1

Question 2

Question 3