N_{2(g)} + 3H_{2(g)} rightleftharpoons 2NH_{3 | vedclass

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(C) The balanced chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$Calculate the number of moles of reactants:$n(N_{2}) = \fra

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$N_{2}, 1.42 \ moles$

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(C) The balanced chemical equation is: $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$Calculate the number of moles of reactants:$n(N_{2}) = \frac{20 \ g}{28 \ g/mol} \approx 0.714 \ mol$$n(H_{2}) = \frac{5 \ g}{2 \ g/mol} = 2.5 \ mol$Determine the limiting reagent by dividing moles by the stoichiometric coefficient:For $N_{2}: \frac{0.714}{1} = 0.714$For $H_{2}: \frac{2.5}{3} \approx 0.833$Since $0.714 The amount of $NH_{3}$ produced depends on the limiting reagent:$n(NH_{3}) = 2 	imes n(N_{2}) = 2 	imes 0.714 = 1.428 \ mol \approx 1.42 \ mol$.

$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
$20 \ g \quad \quad \quad \quad 5 \ g$
Consider the above reaction. The limiting reagent of the reaction and the number of moles of $NH_{3}$ formed are,respectively:

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$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$$20 \ g \quad \quad \quad \quad 5 \ g$Consider the above reaction. The limiting reagent of the reaction and the number of moles of $NH_{3}$ formed are,respectively: