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Stoichiometry and Stoichiometric Calculations

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


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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