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 3.20 g of butane?

Molar mass of butane = 12g/mol*4+1g/mol*10=58g/mol

Mass of 2 moles of butane=2mol*58g/mol=116g

2c4h10 (g) + 13o2 (g) - >8co2 (g) + 10h2o (l)

116g 8 moles

3.20g x

116g butane/8moles CO2=3.20g butane/x

x=3.20g butane*8moles CO2/116g butane=0.2207moles CO2

T=23°C=296K

PV=nRT

V=nRT/P=0.2207moles*0.082 (atm*dm³ / (K/mol)) * 296K / (1atm) = 5.36dm³