$$60g$$ $$CH3COOH$$ & $$23g$$ $$C2H5OH$$ are mixed & heated until equilibrium is reached. If $$4.5g$$ $$H2O$$ is present at equilibrium, total number of moles at equilibrium is

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$$60g$$     $$CH3COOH$$     & $$23g$$     $$C2H5OH$$     are mixed & heated until equilibrium is reached. If $$4.5g$$     $$H2O$$     is present at equilibrium, total number of moles at  equilibrium is

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$$CH3COOH$$  $$+$$  $$C2H5OH$$  ⇌  $$CH3COOC2H5+H2O$$               $$1$$                              $$12$$                            $$0$$                        $$0$$        (1$$−$$x)                   (12$$−$$x)                       x                        x     Given that    $$x=4.518=14$$     Total moles of equilibrium $$=1$$  −  x  $$+12$$  −  x  $$+$$  $$2x$$                                                      $$=32$$

60g CH3COOH & 23g C2H5OH are mixed & heated until equilibrium is reached. If 4.5g H2O is present at equilibrium, total number of moles at equilibrium is

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