Determine the theoretical yield and the percent yield if 21.8 g of K₂CO₃ is produced from reacting 27.9 g KO₂ with 29.0 L of CO₂ (at STP). The molar mass of KO₂ is 71.10 g mol^-1 and K₂CO₃is 138.21 g mol^-1. 4KO₂(s) + 2CO₂(g) → 2K₂CO₃(s) + 3O₂(g) A) 27.1 g, 80.4% yield B) 179 g, 12.2% yield C) 91.7 g, 23.8% yield D) 206 g, 10.6% yield E) 61.0 g, 35.7% yield

Question

Quizplus · Accepted Answer

The balanced chemical equation is:4KO2(s) + 2CO2(g) → 2K2CO3(s) + 3O2(g)The molar mass of KO2 is 71.10 g/mol, and the molar mass of K2CO3 is 138.21 g/mol.The number of moles of KO2 used is:27.9 g KO2 × (1 mol KO2 / 71.10 g KO2) = 0.392 mol KO2The number of moles of CO2 used is:29.0 L CO2 × (1 mol CO2 / 22.4 L CO2) = 1.29 mol CO2The limiting reactant is KO2, because it produces the least number of moles of product.The theoretical yield of K2CO3 is:0.392 mol KO2 × (2 mol K2CO3 / 4 mol KO2) × (138.21 g K2CO3 / 1 mol K2CO3) = 27.1 g K2CO3The percent yield is:(actual yield / theoretical yield) × 100% = (21.8 g / 27.1 g) × 100% = 80.4%Therefore, the correct answer is A: 27.1 g, 80.4% yield.

Determine the Theoretical Yield and the Percent Yield If 21