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%

just did the assignment and the answer was #1 +2 n

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error