Question

Find out the amount of water the will formed when 3.975 gm. of pure cupric oxide are reduced by pure hydrogen also find loss of weight of the cupric oxide

Answer 1

Answer:

Explanation:

According to reaction

CuO (s) + H2(g) ==> Cu(s) + H2O(l)

1mol of CuO gives => 1 mol H2O

i.e

79.5 gm CuO gives => 18gm H2O

Now assuming amount of water formed to be x gm

79.5 gm CuO gives => 18gm H2O

3.975 gm CuO ==> x gm H2O

Cross multiplying..!

79.5× x = 3.975 × 18

x = (3.975 × 18) / 79.5

Therefore x = 0.9 grams

Therefore amount of water formed from 3.975 gm of CuO is [0.9 grams]

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