Question

Two gases in adjoining vessels were brought into correct by opening a stopcock between them. The one vessel measured `0.25 L` and contained `NO` gas at `800` torr and `220 K`, the other measured `0.1 L` contained `O_(2)` gas at `600` torr and `220 K`. The reaction to form `N_(2)O_(4)(s)` exhausts the limiting reagent completely,
(`a`) Neglacting the vapour pressure of `N_(2)O`, what is the pressure of the gas remaining at `220 K` after complection of the reaction?
(`b`) What weight of `N_(2)O` is formed?

Answer 1

`n_(NO)=(800//760xx0.25)/(0.821xx220)=0.0146`
and
`n_(O_(2))=(600//760xx0.1)/(0.0821xx220)=4.37xx10^(-3)`
moles of `NO` used `=2xx4.37xx10^(-3)`
moles of `NO` left `=0.0146-2xx4.37xx10^(-3)`
`=5.86xx10^(-3)`
Final volume `=0.1+0.25=0.35 L`
`P` due to `NO` left `=(5.86xx10^(-3))/(0.35)xx0.0821xx220=0.3 atm`
Now, find mass of `N_(2)O_(4)`
`1 mol` of `O_(2)=1 mol` of `N_(2)O_(4)=4.37xx10^(-3)`
`=92xx4.37xx10^(-3)=0.402 g`