1gm mixture of cuo and cu2o was quantitatively reduced to 0.85gm of metallic copper what was the weight of cuo and cu2o in original sample

Let the amount of cupric oxide in the mixture be W gm. Then, the amount of cuprous oxide in the mixture is (1- W) gm.

Molar mass of copper (Cu) = 63.55 g

Molar mass of cupric oxide (CuO) = 63.55 + 16 = 79.55 g

Molar mass of cuprous oxide (Cu2O) = (63.55 x 2) + 16 = 127.1 + 16 = 143.1 g

1 mole of CuO (79.55 g) on reduction (removal of oxygen) gives 1 mole (63.55 g) of copper.

Amount of copper produced from W g of CuO = (63.55 x W) /79.55 g …………….(1)

1 mole of Cu2O (143.1 g) of Cu2O on reduction gives 2 moles (63.55 x 2 g) of copper.

Amount of copper produced from (1-W) g of Cu2O = 127.1 x (1-W)/ 143.1 ……….(2)

Adding (1) and (2), total amount of copper produced = [(63.55 x W)/79.55] + [127.1 x (1-W)/143.1] g

But the total amount of copper produced by reduction of the mixture = 0.85 g (given)

So, [(63.5 x W)/79.55] + [127.1 x (1-W)/143.1] = O.85

On simplification, W = 0.43

So, the weight of cupric oxide (CuO) in the sample = 0.43 g (answer)

And the weight of cuprous oxide (Cu2O) in the sample = 1– 0.43 = 0.57 g (answer)