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

`{:(,2H_(2)(g)+O_(2)(g)rarr2H_(2)O(g)),("Initially moles"," 2 1 0"),("Moles at equiv.",2-1.6" "1-0.8" "1.6),(,=0.4" "=0.2):}`
Total moles after reaction `=0.4+0.2+1.6=2.2`
Let the initial pressure at `20^(@)C=P_(1)`
Pressure at `120^(@)C=P_(2)`
According to Gay - Lussac.s Law
`(p_(1))/(p_(2))=(T_(1))/(T_(2)) or (0.8)/(p_(2))=(293)/(393)`
`"or "p_(2)=(0.8xx393)/(293)="1.073 atm."`
`"Now, "("Initial moles")/("Moles after equiv.")=("Initial pressure")/("Pressure after equive.")`
`(3)/(2.2)=(1.073)/(p_("equive"))`
`"or "p_("equive")=(1.073xx2.2)/(3)="0.787 atm."`