LAQ : Find out the amount of water that will be formed when 3.975 gm. of pure cupric oxide is reduced by pure hydrogen

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.