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?

Answer 1

Molar mass of C4H10 = 4*MM(C) + 10*MM(H)

= 4*12.01 + 10*1.008

= 58.12 g/mol

mass of C4H10 = 1.6 g

mol of C4H10 = (mass)/(molar mass)

= 1.6/58.12

= 0.0275 mol

From balanced chemical reaction, we see that

when 2 mol of C4H10 reacts, 8 mol of CO2 is formed

mol of CO2 formed = (8/2)* moles of C4H10

= (8/2)*0.0275

= 0.1101 mol

we have:

P = 1.0 atm

n = 0.1101 mol

T = 23.0 oC

= (23.0+273) K

= 296 K

we have below equation to be used:

P * V = n*R*T

1 atm * V = 0.1101 mol* 0.0821 atm.L/mol.K * 296 K

V = 2.68 L

Answer: 2.68 L