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.

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