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

Question

2C4H10(g)+13O(g)-> 8CO2(g)+10H2O(l)

At 1.00 atm and 23C , what is the volume of carbon dioxide formed by the combustion of 3.40 grams of butane?

PV=nRT or n=PV/RT
R=0.08206

Answer 1

You made a typo on the equation. You omitted the 2 on O2.
mols butane = 3.40g/molar mass C4H10 = ?
Convert mols C4H10 to mols CO2 using the coefficients in the balanced equation.
Use PV = nRT to convert mols CO2 to volume in L.

Answer 2

Thanks so much!