The question is incomplete, the complete question is;
.On heating 12.4 g of copper (II) carbonate in a crucible only 7.0g of copper (II) oxide was produced. What was the % yield of copper (II) oxide ? [Cu=64,C=12,O=16]
Answer:
87.5%
Explanation:
CuCO3 -------> CuO + CO2
Number of moles of CuCO3 = 12.4g/123.55 g/mol = 0.1 moles
Since the reaction is 1:1,
Mass of CuO produced = 0.1 moles × 80g/mol = 8 g
Hence,
% yield = actual yield/theoretical yield × 100
% yield = 7/8 × 100
% yield = 87.5%
