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$60 g$ $C H_{3} C O O H$ & $23 g$ $C_{2} H_{5} O H$ are mixed & heated until equilibrium is reached. If $4.5 g$ $H_{2} O$ is present at equilibrium, total number of moles at equilibrium is

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$\frac{3}{2}$

4.5

answer is C.

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Detailed Solution

$C H_{3} C O O H + C_{2} H_{5} O H ⇌ C H_{3} C O O C_{2} H_{5} + H_{2} O$

$1$ $\frac{1}{2}$ $0$ $0$

$(1 - x)$ $(\frac{1}{2} - x)$ $x$ $x$

Given that $x = \frac{4.5}{18} = \frac{1}{4}$

Total moles of equilibrium $= 1 - x + \frac{1}{2} - x + 2 x$

$= \frac{3}{2}$

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