Question

Question #934b0

Answer 1

We need (i) a stoichiometrically balanced equation:

`C_4H_10 + 13/2O_2 rarr 4CO_2(g) + 5H_2O(g)`

Explanation:

And then (ii) equiv quantities of reactant and product:

`"Moles of butane"` `=` `(5.80*g)/(58.12*g*mol^-1)=0.0998*mol`

`"Moles of dioxygen"=(1*atmxx15.0*L)/(0.0821*L*atm*K^-1*mol^-1xx273.15K)=0.669*mol`

(This is not quite `"STP"`, but you will have to adapt this expression.) At any rate there is stoichiometric dioxygen for complete combustion, which we must presume.

Given the equation (i), `4xx0.0998*mol` `CO_2` are evolved per equiv butane.

`V=(nRT)/P`

`=(4xx0.0998*molxx(0.0821*L*atm)/(K*mol)xx273.15K)/(1*atm)=8.95L`

And thus approx. `9*L` carbon dioxide gas evolve from complete combustion.