Question

Liquid octane CH3CH26CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . What is the theoretical yield of carbon dioxide formed from the reaction of 2.28g of octane and 14.2g of oxygen gas? Be sure your answer has the correct number of significant digits in it.

Answer 1

Molar mass of C3H8,

MM = 3*MM(C) + 8*MM(H)

= 3*12.01 + 8*1.008

= 44.094 g/mol

mass(C3H8)= 2.28 g

number of mol of C3H8,

n = mass of C3H8/molar mass of C3H8

=(2.28 g)/(44.094 g/mol)

= 5.171*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 14.2 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(14.2 g)/(32 g/mol)

= 0.4437 mol

Balanced chemical equation is:

C3H8 + 5 O2 ---> 3 CO2 + 4 H2O

1 mol of C3H8 reacts with 5 mol of O2

for 0.051708 mol of C3H8, 0.258539 mol of O2 is required

But we have 0.44375 mol of O2

so, C3H8 is limiting reagent

we will use C3H8 in further calculation

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

According to balanced equation

mol of CO2 formed = (3/1)* moles of C3H8

= (3/1)*0.051708

= 0.155123 mol

mass of CO2 = number of mol * molar mass

= 0.1551*44.01

= 6.83 g

Answer: 6.83 g