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
