Question

When 22.4 litres of H₂(g) is mixed with 11.2 litres of Cl₂(g), each at S.T.P., the moles of HCl(g) formed is equal to :
(a) 1 mole of HCl(g)
(b) 2 moles of HCl(g)
(c) 0.5 moles of HCl(g)
(d) 1.5 moles of HCl(g)

Answer 1

Answer:

At STP, 1 mole of any gas occupies a volume of 22.4 L.

22.4lt

H

2

+

11.2lt

Cl

2

→2HCl

Limiting reagent is Cl

2

.

1 mole of chlorine forms 2 moles of HCl.

Hence, 0.5 moles of chlorine will form 1 mole of HCl.

Thus, when 22.4 liters of H

2

(g) is mixed with 11.2 liters of Cl

2

(g), each at STP, the moles of HCl(g) formed is equal to 1 mol of HCl(g).

Explanation:

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