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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 :-

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Solution

nH2=V(L)22.4=22.422.4=1 mol

nCl2=11.222.4=0.5 mol

H2(g)+Cl2(g)2HCl(g)

initially - 1 mol 0.5 mol 0

Here, Cl2 is the limiting reagent. Thus Cl2 will decide the moles of HCl formed.

Moles of HCl formed =21×0.5=1 mol

Hence, the correct answer is option (a).

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