In the lab, 11.20 g of potassium hydroxide reacted with excess calcium chloride to produce 7.00 g of calcium hydroxide.
2KOH + CaCl2 → 2KCl + Ca(OH)2
a) What is the actual yield of calcium hydroxide produced in the reaction?
All of your calculations need to be shown in order to get full marks on this question.
b) What is the theoretical yield of calcium hydroxide based on the balanced equation for the reaction?
c) What is the percentage yield of the reaction?
a) Since 7.00 g of calcium hydroxide were produced in the lab, the actual yield equals 7.00 g;
b) According to the reaction, the theoretical yield of calcium hydroxide equals:
m(Ca(OH)2) = m(KOH) × Mr(Ca(OH)2) / (Mr(KOH) × 2) = 11.20 g × 74.09 g/mol / (56.11 g × 2) = 7.4 g;
c) The percentage yield of the reaction:
w% = (actual yield / theoretical yield) × 100 % = (7.00 g / 7.4 g) × 100% = 94.6%.