copper II carbonate in a crucible produced only 7g of copper II oxide. What was the percentage yield of copper II oxide​

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]

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%