Question

What volume oi )at S.TP. will be produced on healing, 65.6 g of Ca (NO;)?

Answer 1

The Reaction Equation is as follows:

2Ca(NO3)2-----------> 2CaO + 4NO2 + O2

65.6 g of of Ca(NO3)2 is used.

R.M.M of Ca(NO3)2 is 40+2*14+6*16 => 40+28+96

=>164 g mol-1

So, 65.6 g of Ca(NO3)2 is 65.6/164 mol.

65.6 g of Ca(NO3)2 is 0.4 mol.

So, 2 parts is 0.4 mol.

1 part is 0.4/2 =>0.2 mol

As, 1 part of O2 is produced and 4 parts of NO2 are produced.

0.2 mol of O2 and 0.2*4=0.8 mol of NO2 are produced respectively.

As, at stp 1 mol of any gas has volume of 22.4 litres.

0.2*22.4 litres of O2 and 0.8*22.4 litres of NO2 are evolved respectively.

4.48 litres of O2 and 17.92 litres of NO2 are formed respectively.