Question

Suppose 0.420 kg of octane (C8H18) is burned in air at a pressure of 1 atm and temperature of 20°C, what volume of carbon dioxide gas is produced?

Answer 1

Molar mass of C8H18 = 8*MM(C) + 18*MM(H)

= 8*12.01 + 18*1.008

= 114.224 g/mol

mass of C8H18 = 420.0 g

we have below equation to be used:

number of mol of C8H18,

n = mass of C8H18/molar mass of C8H18

=(420.0 g)/(114.224 g/mol)

= 3.677 mol

we have the Balanced chemical equation as:

2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O

From balanced chemical reaction, we see that

when 2 mol of C8H18 reacts, 16 mol of CO2 is formed

mol of CO2 formed = (16/2)* moles of C8H18

= (16/2)*3.677

= 29.4 mol

we have:

P = 1.0 atm

n = 29.4 mol

T = 20.0 oC

= (20.0+273) K

= 293 K

we have below equation to be used:

P * V = n*R*T

1 atm * V = 29.4 mol* 0.0821 atm.L/mol.K * 293 K

V = 707 L

Answer: 707 L