A partly roasted copper matte contains 3.00 mol of Cu2O and 2.00 mol of Cu2S, which undergo self-reduction, 2Cu2O + Cu2S -> 6Cu + SO2, with no external reductant. Taking Cu = 63.5 and S = 32, the mass of copper formed and the mass of unreacted solid that remains are

571.5 g of Cu; 79.5 g of Cu2S remains

285.8 g of Cu; 79.5 g of Cu2S remains

571.5 g of Cu; 79.5 g of Cu2S remains

762.0 g of Cu; 143.0 g of Cu2O remains

571.5 g of Cu; 159.0 g of Cu2S remains

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JEE Advanced levelSingle correct optionDifficulty 3 of 3mt210

A partly roasted copper matte contains 3.00 mol of $Cu X_{2} O$ and 2.00 mol of $Cu X_{2} S$ , which undergo self-reduction, $2 Cu X_{2} O + Cu X_{2} S 6 Cu + SO X_{2}$ , with no external reductant. Taking Cu = 63.5 and S = 32, the mass of copper formed and the mass of unreacted solid that remains are

A285.8 g of Cu; 79.5 g of $Cu X_{2} S$ remains
B571.5 g of Cu; 79.5 g of $Cu X_{2} S$ remains
C762.0 g of Cu; 143.0 g of $Cu X_{2} O$ remains
D571.5 g of Cu; 159.0 g of $Cu X_{2} S$ remains

Answer: B

The ratio needed is

Cu X_{2} O : Cu X_{2} S = 2 : 1

. 3.00 mol Cu2O needs 1.50 mol Cu2S, so Cu2O is limiting. Cu formed = 3 x 3.00 = 9.00 mol = 571.5 g. Cu2S left = 2.00 - 1.50 = 0.50 mol x 159 g/mol = 79.5 g.

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A partly roasted copper matte contains 3.00 mol of CuX2​O and 2.00 mol of CuX2​S, which undergo self-reduction, 2CuX2​O+CuX2​S​6Cu+SOX2​, with no external reductant. Taking Cu = 63.5 and S = 32, the mass of copper formed and the mass of unreacted solid that remains are

Answer: B

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