40% of a mixture of 0.2 mol of N_2 and 0.6 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The reaction is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$Initial moles: $n(N_2) = 0.2 \ mol$,$n(H_2) = 0.6 \ mol$,$n(NH_3) = 0 \ mol$.

Vedclass · Accepted Answer

$4:5$

Vedclass · Answer

(A) The reaction is: $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$Initial moles: $n(N_2) = 0.2 \ mol$,$n(H_2) = 0.6 \ mol$,$n(NH_3) = 0 \ mol$.Total initial moles $(n_i) = 0.2 + 0.6 = 0.8 \ mol$.Since $40\%$ of $N_2$ reacts,the amount of $N_2$ reacted is $x = 0.40 	imes 0.2 = 0.08 \ mol$.According to the stoichiometry:$N_2$ reacted $= 0.08 \ mol$$H_2$ reacted $= 3 	imes 0.08 = 0.24 \ mol$$NH_3$ produced $= 2 	imes 0.08 = 0.16 \ mol$Moles at final state:$n(N_2) = 0.2 - 0.08 = 0.12 \ mol$$n(H_2) = 0.6 - 0.24 = 0.36 \ mol$$n(NH_3) = 0.16 \ mol$Total final moles $(n_f) = 0.12 + 0.36 + 0.16 = 0.64 \ mol$.At constant temperature and pressure,$V \propto n$,so the ratio of volumes is the ratio of total moles:Ratio $= \frac{V_f}{V_i} = \frac{n_f}{n_i} = \frac{0.64}{0.8} = \frac{64}{80} = \frac{4}{5}$.

$40\%$ of a mixture of $0.2 \ mol$ of $N_2$ and $0.6 \ mol$ of $H_2$ reacts to give $NH_3$ according to the equation:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
at constant temperature and pressure. What is the ratio of the final volume to the initial volume of gases?

Similar Questions

One mole of fluorine is reacted with two moles of hot concentrated $KOH$. The products formed are $KF, H_2O$ and $O_2$. The molar ratio of $KF, H_2O$ and $O_2$,respectively is

Number of moles of methane required to produce $22 \ g$ $CO_{2(g)}$ after combustion is $x \times 10^{-2}$ moles. The value of $x$ is

Volume of $0.6\, M\, NaOH$ required to neutralize $30\, cm^{3}$ of $0.4\, M\, HCl$ is ............. $cm^{3}$

When $10 \ mol$ of $SO_2$ and $15 \ mol$ of $O_2$ are mixed and passed over a catalyst,$8 \ mol$ of $SO_3$ is formed. How many moles of $SO_2$ and $O_2$ remain unreacted?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$40\%$ of a mixture of $0.2 \ mol$ of $N_2$ and $0.6 \ mol$ of $H_2$ reacts to give $NH_3$ according to the equation:$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$at constant temperature and pressure. What is the ratio of the final volume to the initial volume of gases?