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$22.4 {LC}_{2} H_{6}$ gas at 0° C and 1 atm is burnt in excess of $O_{2}$ gas. The products are passed through 5 L of 2 M KOH solution. The maximum moles of $K_{2} {CO}_{3}$ which may form in the solution is :

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answer is 2.

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Detailed Solution

The complete reaction takes place as,

$\underset{1 moles}{C_{2} H_{6}} + \frac{7}{2} O_{2} ⟶ \underset{}{\underset{2 moles}{2 {CO}_{2}} + 3 H_{2} O}$

$\begin{array}{ccc} \underset{\underline{2}}{{CO}_{2}} + \underset{\underset{6}{10}}{2 KOH} & ⟶ \underset{\underset{2}{-}}{K_{2} {CO}_{3}} + \underline{H_{2} O} \end{array}$

Thus, the maximum moles of $K_{2} {CO}_{3}$ which may form in the solution is 2.

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