Question

When 22.4 litres of H2 (g) is mixed with 11.2 litres of Cl2 (g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to
(a) 2 moles of HCl (g)
(b) 0.5 moles of HCl (g)
(c) 1.5 moles of HCl (g)
(d) 1 moles of HCl (g)

Answer 1

Correct Option is d: 1 mole of HCl

********************************************
Given Chemical equation:
H2(g) +Cl2(g) ---------> 2HCl(g)

1 molecues of H2 reacts with 1 molecules of Cl2 to give 2 molecules of HCl.
H2 +Cl2 ---------> 2HCl

2u     71u               73u  --------- molecular mass

2g        71g            73g

1 mole of H2 reacts with  1mole  of Cl2 to give  2moles of HCl

1 mole of any gas at STP occupies 22.4 litres.
so,
22.4 litres of H2+22.4 litres of Cl2 ---> 2x22.4 litres of HCl


According to given :

22.4 litres of H2 reacts with 11.2 litres of Cl2
ll.2 litres of chlorine=0.5 moles.
As 1 mole of H2 should react with 1mole of Cl2 . But supplied is 0.5.....
so Limiting Reactant is Cl2 and Excess reactant --H2

so production of HCl depends upon Cl2
1mole of Cl2------> 2 moles of HCl
o.5 mole of Cl2 ----->? moles
=0.5x2/1
=1 mole of HCl