Question

15. Consider the following combustion reaction of propane. 3C3H8 + 5O2 → 3CO2 + 4H2O

a) What mass of O2 , in g, would be needed to react with 0.421 kg of C3H8?

b) What mass of CO2 , in g, would be produced from the combustion of 0.421 kg of C3H8 with excess oxygen?

Answer 1

A)
mass of C3H8 = 0.421 kg = 421 g
molar mass of C3H8 = 44.094 g/mol
mol of C3H8 = (mass)/(molar mass)
= 421/44.094
= 9.547784 mol


According to balanced equation
mol of O2 required = (5/3)* moles of C3H8
= (5/3)*9.547784
= 15.912974 mol

mass of O2 = number of mol * molar mass
= 15.912974*32
= 509 g
= 0.509 Kg
Answer: 0.509 Kg

B)

According to balanced equation
mol of CO2 formed = moles of C3H8
= 9.55 mol



mass of CO2 = number of mol * molar mass
= 9.55*44.01
= 420 g
= 0.420 Kg
Answer: 0.420 Kg