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Stoichiometry and Stoichiometric Calculations

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By Expert Faculty of Sri Chaitanya
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

Moderate
Solution

1 mole 22.4 litres at S.T.P. 

nH2=22.422.4=1 mol,nCl2=11.222.4=0.5 mol

reaction 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=1

Here, Cl2 is limiting reagent. So, 1 mole of HCl(g) is formed.


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