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

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Solution

1 mole = 22.4 liters at S.T.P.

$^{n} H_{2} = \frac{22.4}{22.4} = 1 m o l;^{n} C l_{2} = \frac{11.2}{22.4} = 0.5 m o l$

Reaction is as,

$\begin{matrix} H_{2 (g)} & + & C l_{2 (g)} & ⟶ & 2 H C l_{(g)} Initial & 1 m o l & 0.5 m o l & 0 Final & (1 - 0.5) & (0.5 - 0.5) & 2 \times 0.5 = 0.5 m o l & = 0 m o l & 1 m o l \end{matrix}$

Here, $C l_{2}$ 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 $STP$ , the moles of $HCl (g)$ formed are equal to

The number of moles present in 11.2 litres of ${CO}_{2}$ at S.T.P is:

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