Question

When sodium nitrate is heated, the following reaction occurs,

2NaNO3 (s) 2NaNO2(s) + O2(g)

What is the volume of oxygen at STP obtained when 4.25g of NaNO3is heated? (1 mole gas occupies 22.4 dm3at STP)

Answer 1

Moles of NaNO3​ taken =854.25​=0.05 mole

2 moles of Sodium nitrate produces =1 mole oxygen

Then,

0.05 mole sodium nitrate will produce = 0.025 mole oxygen,

So,

Volume of 1 mole oxygen = 22.4 litre

Volume of 0.025 mole oxygen = 0.56 litre = 560 ml $O_2$