Q.

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

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By Expert Faculty of Sri Chaitanya

answer is 1.

(Detailed Solution Below)

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Detailed Solution

1 mole ≡22.4 litres at S.T.P. nH2=22.422.4=1 mol,nCl2=11.222.4=0.5 molreaction is as H2(g) + Cl2(g) ⟶ 2HCl(g) Initial 1 mol0.5 mol 0 Final (1-0.5)=0.5 (0.5-0.5)=0 2×0.5=1Here, Cl2 is limiting reagent. So, 1 mole of HCl(g) is formed.

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