$60g$ $C{H}_{3}COOH$ & $23g$ $C_2{H}_{5}OH$ are mixed & heated until equilibrium is reached. If $4.5g$ $H_{2}O$ is present at equilibrium, total number of moles at equilibrium is

$C{H}_{3}COOH+C_2{H}_{5}OH\rightleftharpoons C{H}_{3}COO{C}_2{H}_5+H_{2}O$

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

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

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

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

                                                $=\frac{3}{2}$