A mixture in which the mole ratio of `H_(2)` and `O_(2)` is `2:1` is used to prepare water by the reaction.
`2H_(2(g))+O_(2(g))rarr2H_(2)O_((g))`
The total pressure in the container is `0.8 atm` at `20^(@)C` before the reaction. Determine the final pressure at `120^(@)C` after reaction assuming `80%` yield of water.

Explanation

The given reaction is
`2H_(2)(g)+O_(2)(g) to 2H_(2)O_(g)`
`{:(,"Initial moles",2n,n,0),(,"Final moles",(2n-2x),(n-x),2x):}`
% yields=80
`therefore (2x)/(2n)xx100=80`
After the reaction,
Number of moles of `H_(2)=2n-2xx0.8n=0.4n`
Number of moles of `O_(2)=0.2n`
Number of moles of `H_(2)O=1.6n`
total moles=`0.4n+0.2n+1.6n=2.2n`
Initial state: pV=nRT
`0.8xxV=3mxxRxx293...(i)`
After the reaction, `PxxV=2.2nxxRxx393...(ii)`
Solving eqs. (i) and (ii), we get
P=0.787atm