For the reaction: N_{2(g)} + 3H_{2(g)} righta | vedclass

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(A) $1$. Calculate moles of reactants: Moles of $N_2 = \frac{168 \ 	ext{g}}{28 \ 	ext{g/mol}} = 6 \ 	ext{mol}$. Moles of $H_2 = \frac{30 \ 	ext{g}

Vedclass · Accepted Answer

$-106$

Vedclass · Answer

(A) $1$. Calculate moles of reactants: Moles of $N_2 = \frac{168 \ 	ext{g}}{28 \ 	ext{g/mol}} = 6 \ 	ext{mol}$. Moles of $H_2 = \frac{30 \ 	ext{g}}{2 \ 	ext{g/mol}} = 15 \ 	ext{mol}$.$2$. Identify limiting reagent: According to the stoichiometry $N_2 + 3H_2 ightarrow 2NH_3$,$1 \ 	ext{mol}$ of $N_2$ requires $3 \ 	ext{mol}$ of $H_2$. For $6 \ 	ext{mol}$ of $N_2$,we need $18 \ 	ext{mol}$ of $H_2$. Since we only have $15 \ 	ext{mol}$ of $H_2$,$H_2$ is the limiting reagent.$3$. Calculate moles of $NH_3$ produced: $3 \ 	ext{mol}$ of $H_2$ produce $2 \ 	ext{mol}$ of $NH_3$. Thus,$15 \ 	ext{mol}$ of $H_2$ produce $\frac{2}{3} 	imes 15 = 10 \ 	ext{mol}$ of $NH_3$.$4$. Calculate $\Delta n_g$: $\Delta n_g = \sum n_p - \sum n_r = 10 - (5 + 15) = -10 \ 	ext{mol}$ for the actual reaction occurring.$5$. Calculate $\Delta U$: $\Delta U = \Delta H - \Delta n_g RT$. Given $\Delta H$ for $2 \ 	ext{mol}$ of $NH_3$ is $-24 \ 	ext{kcal}$,for $10 \ 	ext{mol}$ it is $-24 	imes 5 = -120 \ 	ext{kcal}$.$6$. $\Delta U = -120 - (-10 	imes 0.002 \ 	ext{kcal/mol K} 	imes 700 \ 	ext{K}) = -120 + 14 = -106 \ 	ext{kcal}$.

Column - $A$	Column - $B$
$(A)$ $88 \ g$ of $CO_2$	$(1)$ $0.2 \ mol$
$(B)$ $6.022 \times 10^{23}$ molecules of water	$(2)$ $2 \ mol$
$(C)$ $5.6 \ L$ of $O_2$ at $STP$	$(3)$ $1 \ mol$
$(D)$ $96 \ g$ of $O_2$	$(4)$ $6.022 \times 10^{23}$ molecules
$(E)$ One mole of any gas at $STP$	$(5)$ $3 \ mol$

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