Stoichiometry and Stoichiometric Calculations

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Question

When 22.4 litres of $H_{2} (g)$ is mixed with 11.2 litres of ${Cl}_{2} (g)$ , each at S.T.P, the moles of HCl (g)formed is equal to

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Solution

$1 mole \equiv 22.4 litres at S.T.P.$
$n_{H_{2}} = \frac{22.4}{22.4} = 1 mol, n_{{Cl}_{2}} = \frac{11.2}{22.4} = 0.5 mol$
reaction is as
$H_{2 (g)} + {Cl}_{2 (g)} ⟶ 2 {HCl}_{(g)}$
$\begin{array}{l} Initial & 1 mol & 0.5 mol & 0 \\ Final & (1 - 0.5) = 0.5 & (0.5 - 0.5) = 0 & 2 \times 0.5 = 1 \end{array}$
Here, ${Cl}_{2}$ is limiting reagent. So, 1 mole of HCl(g) is formed.

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Stoichiometry and Stoichiometric Calculations

Unlock the full solution & master the concept.

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 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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When 22.4 litres of $H_{2} (g)$ is mixed with 11.2 litres of ${Cl}_{2} (g)$ , each at S.T.P, the moles of HCl (g)formed is equal to