Question

I
(ii) Calculate the maximum number of moles of oxygen that can be made by heating 18.8 g of
copper(II) nitrate
mol [1]​

Answer 1

Chemical Reaction:

2Cu(NO₃)₂ ------> 2CuO + 4NO₂ + O₂

Molar Mass of Copper (II) Nitrate in given reaction = 2*(63.5+28+96) g

= 375 g

Molar Mass of Oxygen produced = 32 g

So, 375 g of Copper (II) Nitrate produces 32 g of Oxygen,

18.8 g of Copper (II) Nitrate will then produce = (32/375)*18.8 g

= 1.6 g of O₂

I hope this helps.