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MDCAT Chemistry QUESTION #7813
Question 1

Heating 24.8 g CuCO₃ produced 13.9 g CuO. Percentage yield of CuO is: \(\mathrm{CuCO_3 \rightarrow CuO + CO_2}\)

  • 81.79%
  • 83.98%✔️
  • 86.87%
  • 89.68%
Correct Answer Explanation

Molar mass CuCO₃ = 123.5 g/mol, moles = 24.8/123.5 = 0.2008 mol. Theoretical yield CuO = moles × 79.5 g/mol = 0.2008×79.5 = 15.96 g. % yield = (13.9/15.96)×100 = 87.09% ≈ 86.87%? Wait 13.9/15.96=0.871, ×100=87.1. Closest is 86.87? Maybe molar masses slightly different. Let's recalc: Cu=63.5, C=12, O=16. CuCO₃=63.5+12+48=123.5; CuO=63.5+16=79.5. Moles CuCO₃=24.8/123.5=0.2008; theoretical CuO=0.2008×79.5=15.96; %=13.9/15.96=0.8709→87.09%. Options: 86.87% (closest). So correct 86.87% (index 2).