Question

Butane, C4H10 , is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is

2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l)

At 1.00 atm and 23 ∘C , what is the volume of carbon dioxide formed by the combustion of 1.60 g of butane?

Express your answer with the appropriate units.

volume of CO2 =

Answer 1

2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l)

2 moles of C4H10 on combustion to gives 8 moles of CO2

2*58g of C4H10 on combustion to gives 8 moles of CO2

1.6g of C4H10 on combustion to gives = 8*1.6/2*58 = 0.11 moles of CO2

PV = nRT

n   = 0.11moles

R   = 0.0821L-atm/mole-K

P = 1atm

T   = 23C⁰   = 23+273 = 296K

PV = nRT

V   = nRT/P

     = 0.11*0.0821*296/1   = 2.673L >>>>answer

volume of Co2 = 2.673L