Consider the following reaction equilibrium:N | vedclass

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(C) The reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.Let the initial moles be $n(N_2) = 1$,$n(H_2) = 3$,and $n(NH_3) = 0$.At equil

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$3.2$

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(C) The reaction is $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$.Let the initial moles be $n(N_2) = 1$,$n(H_2) = 3$,and $n(NH_3) = 0$.At equilibrium,let $x$ be the extent of reaction:$n(N_2) = 1 - x$$n(H_2) = 3 - 3x$$n(NH_3) = 2x$Given that at equilibrium,$n(N_2) = 0.6$.So,$1 - x = 0.6 \implies x = 0.4$.Now,calculate the moles of each component at equilibrium:$n(N_2) = 0.6 	ext{ mol}$$n(H_2) = 3 - 3(0.4) = 3 - 1.2 = 1.8 	ext{ mol}$$n(NH_3) = 2(0.4) = 0.8 	ext{ mol}$The total number of moles at equilibrium is the sum of the moles of all gases:$	ext{Total moles} = n(N_2) + n(H_2) + n(NH_3) = 0.6 + 1.8 + 0.8 = 3.2 	ext{ mol}$.

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