Answer to Question #70924 in General Chemistry for Desai

Question

If you burn a 40L tank of C8H18 how much CO2 will be produced in kilograms? The density of C8H18 is 703g/L

Answer 1

The combustion reaction of hydrocarbon(C8H18) is :
С_8 H_18+25/2 O_2→8CO_2+9H_2 O
Mr (C8H18)=114 g/mol
Mr (CO2) = 44 g/mol

Density=Mass/Volume,so Mass=Density*Volume

m (C8H18) = ρ (g/L)* V (L) = 703 g/L *40 L = 28120 g
n (C8H18) = m (C8H18)/Mr (C8H18)=28120 g / 114 g/mol = 246.67 mol

Reaction mole ratio is C8H18: CO2=1:8, i.e.
(246.67 mol C_8 H_18)/(1 mol C_8 H_18 )=(x mol CO_2)/(8 mol CO_2 )
Solve simple proportion, n (CO2)=x= 246.67 mol * 8 mol=1973.33 mol
m (CO2) = n (CO2)*Mr (CO2) = 1973.33 mol * 44 g/mol = 86828.67 g /1000=86.83 kg

Answer: 86.83 kg