Question
Find out the amount of water that will be formed when 3.975 gm. of pure cupric oxide is reduced by pure hydrogen. Also, find the loss of weight of the cupric oxide. [Cu = 63.5, H = 1, 0 = 16]
Answer
Balanced Chemical Equation :
CuO + H2 = H2O + Cu
The molecular weight of CuO = 63.5 + 16 = 79.5 g
The molecular weight of H2O = 1×2 + 16 = 18 g
Clearly, 79.5 g of CuO produces 18 g of water
Hence, 3.975 g of CuO produces 18×3.975\over 79.5
= 0.9 g of water
Now, from the above reaction
79.5 g of CuO is reduced to 63.5 g of Cu
∴ 3.975 g of CuO is reduced to 63×3.975\over 79.5
= 3.175 g of Cu
∴ Loss of wt = 3.975 – 3.175 = 0.8 g.