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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:
A
1 mol of HCl(g)
B
2 mol of HCl(g)
C
0.5 mol of HCl(g)
D
1.5 mol of HCl(g)
Detailed Solution
$n_{H_2} = \frac{22.4}{22.4} = 1$ mol; $n_{Cl_2} = \frac{11.2}{22.4} = 0.5$ mol
$H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$
$Cl_2$ is the limiting reagent: 0.5 mol $Cl_2$ gives $0.5\times2 = 1$ mol HCl (0.5 mol $H_2$ is left).
$H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$
$Cl_2$ is the limiting reagent: 0.5 mol $Cl_2$ gives $0.5\times2 = 1$ mol HCl (0.5 mol $H_2$ is left).