56 , g of nitrogen and 8 , g of hydrogen gas | vedclass

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Question

(C) The balanced chemical equation is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$Initial moles: $n(N_2) = \frac{56 \, g}{28 \, g/mol} = 2 \, mol$,

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$1, 1, 2$

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(C) The balanced chemical equation is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$Initial moles: $n(N_2) = \frac{56 \, g}{28 \, g/mol} = 2 \, mol$,$n(H_2) = \frac{8 \, g}{2 \, g/mol} = 4 \, mol$,$n(NH_3) = 0 \, mol$.At equilibrium,$34 \, g$ of $NH_3$ is present,which is $n(NH_3) = \frac{34 \, g}{17 \, g/mol} = 2 \, mol$.According to the stoichiometry of the reaction,$2 \, mol$ of $NH_3$ are produced from $1 \, mol$ of $N_2$ and $3 \, mol$ of $H_2$.Equilibrium moles of $N_2 = 2 \, mol - 1 \, mol = 1 \, mol$.Equilibrium moles of $H_2 = 4 \, mol - 3 \, mol = 1 \, mol$.Equilibrium moles of $NH_3 = 2 \, mol$.Thus,the moles are $1, 1, 2$.

$56 \, g$ of nitrogen and $8 \, g$ of hydrogen gas are heated in a closed vessel. At equilibrium,$34 \, g$ of ammonia are present. The equilibrium number of moles of nitrogen,hydrogen,and ammonia are respectively:

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