For the reaction shown, compute the theoretical yield of product (in moles) for each of the following initial amounts of reactants

/ Exams & Certification / By

For the reaction shown, compute the theoretical yield of product (in moles) for each of the following initial amounts of reactants.
Mn(s)+O2(g)?MnO2(s)

Part A 5mol Mn, 5mol O2 Express your answer using one significant figure. __________moll

Part B 3mol Mn, 9mol O2 Express your answer using one significant figure. ___________________moll

Part C 27.0mol Mn, 43.8mol O2 ___________________moll

The Correct Answer and Explanation is :

To determine the theoretical yield of MnO₂ (manganese dioxide) for each case, we need to use stoichiometry based on the balanced chemical equation:

[
2Mn(s) + O_2(g) \rightarrow 2MnO_2(s)
]

Step 1: Identify the Limiting Reactant

  • From the balanced equation, 2 moles of Mn react with 1 mole of O₂ to produce 2 moles of MnO₂.
  • The limiting reactant determines the amount of product formed.

Part A: 5 mol Mn, 5 mol O₂

  1. Using Mn:
  • ( 5 \text{ mol Mn} \times \frac{2 \text{ mol MnO}_2}{2 \text{ mol Mn}} = 5 \text{ mol MnO}_2 )
  1. Using O₂:
  • ( 5 \text{ mol O}_2 \times \frac{2 \text{ mol MnO}_2}{1 \text{ mol O}_2} = 10 \text{ mol MnO}_2 )
  • Since Mn produces the smaller amount of MnO₂ (5 mol), Mn is the limiting reactant.
  • Theoretical yield: 5 mol MnO₂ (1 sig. fig.) → 5 mol

Part B: 3 mol Mn, 9 mol O₂

  1. Using Mn:
  • ( 3 \text{ mol Mn} \times \frac{2 \text{ mol MnO}_2}{2 \text{ mol Mn}} = 3 \text{ mol MnO}_2 )
  1. Using O₂:
  • ( 9 \text{ mol O}_2 \times \frac{2 \text{ mol MnO}_2}{1 \text{ mol O}_2} = 18 \text{ mol MnO}_2 )
  • Since Mn produces the smaller amount of MnO₂ (3 mol), Mn is the limiting reactant.
  • Theoretical yield: 3 mol MnO₂ (1 sig. fig.) → 3 mol

Part C: 27.0 mol Mn, 43.8 mol O₂

  1. Using Mn:
  • ( 27.0 \text{ mol Mn} \times \frac{2 \text{ mol MnO}_2}{2 \text{ mol Mn}} = 27.0 \text{ mol MnO}_2 )
  1. Using O₂:
  • ( 43.8 \text{ mol O}_2 \times \frac{2 \text{ mol MnO}_2}{1 \text{ mol O}_2} = 87.6 \text{ mol MnO}_2 )
  • Since Mn produces the smaller amount of MnO₂ (27.0 mol), Mn is the limiting reactant.
  • Theoretical yield: 27 mol MnO₂ (1 sig. fig.) → 30 mol

Explanation (300 words)

Stoichiometry allows us to determine the theoretical yield of a reaction by using the balanced chemical equation. In this reaction, manganese (Mn) reacts with oxygen gas (O₂) to form manganese dioxide (MnO₂), following a 2:1:2 mole ratio.

For each case, we determine the limiting reactant by comparing the ratio of reactants to the balanced equation. The limiting reactant is the reactant that produces the least amount of product, which in turn determines the theoretical yield.

  • Part A: 5 mol Mn and 5 mol O₂ are given. Since 5 mol Mn requires only 2.5 mol O₂ but we have excess O₂, Mn is the limiting reactant, producing 5 mol MnO₂.
  • Part B: 3 mol Mn and 9 mol O₂ are given. Since 3 mol Mn requires only 1.5 mol O₂ but we have excess O₂, Mn is again the limiting reactant, producing 3 mol MnO₂.
  • Part C: 27.0 mol Mn and 43.8 mol O₂ are given. Since 27.0 mol Mn requires only 13.5 mol O₂ but we have excess O₂, Mn is still the limiting reactant, producing 30 mol MnO₂ (rounded to 1 sig. fig.).

In each case, excess reactants remain unused. The theoretical yield is found using mole ratios, and the answer is reported with one significant figure, as required.

Let me know if you need further clarification! 🚀

Related Posts

Scroll to Top