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),to,2H_(2)O(g)),(" Intial moles" ,2a ,,a,,0):}`
As pressure are in the ratio of their moles
` :. 2 a + a = 0*8 "atm" or 3 a = 0*8 or a = (0*8)/3 "atm " `
Theoretically expected yield of `H_(2)O = 2 a`
Actual yield ` 2 a xx 90/100 = 1*6 a`
` :. H_(2) reacted = 1*6 a " moles ", O_(2) "reacted "= 0* 8 a " moles "`
Moles after reaction : ` H_(2) = 2 a - 1*6 a = 0* 4 a, O_(2) = a - 0*8 a = 0*2 a `
Total no. of moles ` = 0*4 a + 0*2 + 1*6 a = 2*2 a `
Hence, final pressure ` = 2*2 xx (0*8)/3 = 0* 59 "atm"`