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 3.20 g of butane?

Answer 1

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