The 'roasting' of 100.0 g of a copper ore yielded 71.8 g pure copper. If the ore is compound of Cu 2 S and CuS with 4.5% inert impurity, calculate the percent of Cu 2 S in the ore. (Atom

Question

The 'roasting' of 100.0 g of a copper ore yielded 71.8 g pure copper. If the ore is compound of $Cu_{2} S$ and CuS with $4.5%$ inert impurity, calculate the percent of $Cu_{2} S$ in the ore. (Atomic mass of ' Cu ' is 63.5).

The reactions are:

$C u_{2} S + O_{2} \to 2 C u + S O_{2}$

$C u S + O_{2} \to C u + S O_{2}$

TestHub · Accepted Answer

Mass of Cu 2 ​ S & CuS = 100 − 4.5 = 95.5 g Let mass of Cu 2 ​ S is x g . Cu 2 ​ S + O 2 ​ → 2 Cu + SO 2 ​ CuS + O 2 ​ → Cu + SO 4 ​ Mass of Cu from Cu 2 ​ S + Mass of Cu from CuS = 71.8 159 x ​ × 63.3 × 2 + 95.5 ( 96.5 − x ) ​ 63.5 = 71.8 x ( 159 127 ​ − 95.5 63.5 ​ ) = 8.3 x = 0.134 8.3 ​ = 62.01 g ∴ % of Cu 2 ​ S is 62

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