Question

Two gases in adjoining vessels are brought into contact by opening a stop cock between them. The one vessel measured 0.25 lit and contained NO at 800 torr and 220 K, the other measured 0.1 lit and contained $O_2$ at 600 torr. The reaction to form $N_2{O}_4$ (solid) exhausts the limiting reactant completely.
a) Neglecting the vapour pressure of $N_2{O}_4$ what is the pressure of the gas remaining at 220 K after completion of the reaction?
b) What weight of $N_2{O}_4$ is formed?
[Given that : $\frac{342.8}{760}\times \frac{0.35}{0.082}\times \frac{1}{22}=8.75\times 10$]

Answer 1

number of mole of No in 1st vessel

$n_1=\frac{PV}{RT}$

$=\frac{\left(800torr\right)\left(0.25lit\right)}{R\times \left(220k\right)}$

number of mole $O_2$ in 11nd vessel

$n_2=\frac{PV}{RT}$

$=\frac{\left(600torr\right)\left(0.1lit\right)}{R\times \left(220k\right)}$

$2NO+O_2\to N_2{O}_4$

After completion of this reaction $O_2$ is limiting reapet (by comparing $n_1$ and $n_2$)

number of mole of NO left $=n_1-2n_2$

Number of mole of $N_2{O}_4$ formed =$n_2$

(1) pressure $=\frac{(n_1-2n_2)V}{R\times 220k}$

$=\frac{-\frac{1}{R\times 220k}(800\times 0.25-2\times 600\times 0.1)}{R\times 220k}\times 0.35$

$=80torr\times 0.35$

$=28torr$

(2) $n_2=\frac{600\times \frac{1}{760}\times 0.1}{0.0821\times 220}=0.00437$

Mass = $0.00437\times 92=0.40g$