Question #934b0

1 Answer
anor277
Dec 16, 2016

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.

Related questions
See all questions in Gas Stoichiometry
Impact of this question
1862 views around the world
You can reuse this answer
Creative Commons License