Question

How many kilograms of CO₂ would be produced through combustion of a gallon (2.6 kg) of octane (C₈H₁₈)?

2 C₈H₁₈ (g) + 25 O₂ (g) --> 16 CO₂ (g) + 18 H₂O (g)

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 = 2.60 Kg = 2600 g

mol of C8H18 = (mass)/(molar mass)

= 2600/114.224

= 22.7623 mol

we have the Balanced chemical equation as:

2 C8H18 (g) + 25 O2 (g) --> 16 CO2 (g) + 18 H2O (g)

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)*22.7623

= 182.0983 mol

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = number of mol * molar mass

= 182.0983*44.01

= 8014 g

= 8.01 Kg

Answer: 8.01 Kg