Question

A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What is the percent yield?

Answer 1

Answer:

81.26% is the percent yield

Explanation:

Based on the reaction:

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.

To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

Actual yield (0.366g) / Theoretical yield * 100

Moles CaCl₂ = Moles CaCO₃:

0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:

0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

Percent yield = 0.366g / 0.450g * 100

81.26% is the percent yield