Question

Butane `(C_(4)H_(10))` gas burns in oxygen to give carbon dioxide and water accoding to the reaction.
`2C_(4)H_(10) (g) + 13O_(2)(g) rarr 8 CO_(2)(g) + 10H_(2)O(l)`
When `5.0 L` of butane ware burnt ini excess of oxygen at `67^(@)C` and 2 bar pressure, calculate the dioxide evolved.

Answer 1

Correct Answer - `12.62 L`
Step I. Volume of `CO_(2)` at `N.T.P.`
The chemical equation for the reaction is :
`underset(2L)(2C_(4)H_(10)(g))+13O_(2)(g)rarrunderset(8L)(8CO_(2)(g))+10H_(2)O(l)`
2L of butane evolve `CO_(2) = 8L` , 5 L of butane evolve `CO_(2) = 8 xx 5/2 = 20 L`
Step II. Volume of `CO_(2)` under experimental conditions
`{:("N.T.P. conditions",,"Experimental conditions"),(V_(1)=20L,,V_(2)=?),(P_(1)=1.013"bar",,P_(2)=2 "bar"),(T_(1)=273 K,,T_(2)=67+273=300K):}`
By applying gas equation :
`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = ((1.013 "bar") xx (20 L) xx (340 K))/((2 "bar") xx (273)) = 12.62 L`