4 litre `C_(2)H_(4)` (g) burns in oxygen at `27^(@)C` and 1 atm to produce `CO_(2)` (g) and `H_(2)O` (g) Calculate the volume of `CO_(2)` formed at (a) `27^(@)C` and 1 atm (b) `127^(@) C` and 1atm ( c) `27^(@)C` and 2 atm .

Explanation

`C_(2)H_(4) (g) + 3O_(2) g rarr2CO_(2(g)) + 2H_(2)O`
Under same conditions of `P` and `T` volume of gases react in their mole ration and produce the products in the same molar ratio and produce the products in the same molar ratio Thus at `27^(@) C` and 1 atm 1 vol. or 1 mole of `C_(2)H_(4)` gives =2 volume `CO_(2)`
`:. 4` vol of `C_(2)H_(4)` gives = `2 xx 4` volume `CO_(2)`
= 8 litre `CO_(2)`
(b) Now, at `127^(@)C` and 1 atm ` : (P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`
or `(1 xx8)/(300) = (1xxV)/(400)`
`V = 10.67` litre
(c) Similarly at `27^(@)C` and 2atm
`:. V =4` litre .