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 1.01 bar and 23 ∘C, what is the volume of carbon dioxide formed by the combustion of 2.60 g of butane?

Answer 1

Molar mass of C4H10,

MM = 4*MM(C) + 10*MM(H)

= 4*12.01 + 10*1.008

= 58.12 g/mol

mass(C4H10)= 2.60 g

number of mol of C4H10,

n = mass of C4H10/molar mass of C4H10

=(2.6 g)/(58.12 g/mol)

= 4.474*10^-2 mol

From reaction,

moles of CO2 formed = (8/2)*moles of C4H1O reacted

= 4*4.474*10^-2 mol

= 0.17896 mol

Given:

P = 1.01 bar

= (1.01/1.01325) atm

= 0.9968 atm

n = 0.17896 mol

T = 23.0 oC

= (23.0+273) K

= 296 K

use:

P * V = n*R*T

0.9968 atm * V = 0.179 mol* 0.0821 atm.L/mol.K * 296 K

V = 4.36 L

Answer: 4.36 L