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

`p_(Hz) = 2/3 xx 0.8 = 0.533` atm
`pO_(2) = 1/3 xx 0.8 = 0.266` atm
`{:(,2H_(2),+,O_(2),to,2H_(2)O),(t=0,0.533,,0.266,,0),("After reaction",(0.533 xx 20)/100,,(0.266 xx 20)/100,,(0.533 xx 80)/100),(,=0.1066,,=0.0533,,=0.4264):}`
Total pressure `= 0.1066 + 0.0533 + 0.4264 = 0.5863` atm
Using Gay-Lussac.s law, ` = P_(1)/T_(1) = P_(2)/T_(2)`
` = (0.5863)/(293) = P_(2)/393 `
`P_(2) = 0.7864` atm