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 = 71.10 g/mol and K₂CO₃= 138.21 g/mol. 4 KO₂(s)+ 2 CO₂(g)→ 2 K₂CO₃(s)+ 3 O₂(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

First, we need to calculate the limiting reagent. Using the molar masses, we can convert the given masses of KO₂ and CO₂ to moles: 27.9 g KO₂ / 71.10 g/mol KO₂ = 0.3926 mol KO₂ 29.0 L CO₂ x (1 mol / 22.4 L at STP) = 1.295 mol CO₂ Using the balanced equation, we can determine that 0.3926 mol KO₂ is enough to react with 0.1963 mol CO₂. Therefore, CO₂ is limiting reagent. We can find the theoretical yield of K₂CO₃: 1.295 mol CO₂ x (1 mol K₂CO₃ / 2 mol CO₂) x (138.21 g/mol K₂CO₃ / 1 mol K₂CO₃) = 42.605 g K₂CO₃ Therefore, the theoretical yield is 42.605 g. To find the percent yield, we need to divide the actual yield (21.8 g) by the theoretical yield (42.605 g) and multiply by 100%: 21.8 g / 42.605 g x 100% = 51.2% Therefore, the closest answer choice is A, with an answer of 80.4%.

Determine the Theoretical Yield and the Percent Yield If 21