Question

Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 14. g of hexane is mixed with 15.6 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 3 significant digits.

Answer 1

Molar mass of C6H14,

MM = 6*MM(C) + 14*MM(H)

= 6*12.01 + 14*1.008

= 86.172 g/mol

mass(C6H14)= 14.0 g

number of mol of C6H14,

n = mass of C6H14/molar mass of C6H14

=(14.0 g)/(86.172 g/mol)

= 0.1625 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 15.6 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(15.6 g)/(32 g/mol)

= 0.4875 mol

Balanced chemical equation is:

2 C6H14 + 19 O2 ---> 12 CO2 + 14 H2O

2 mol of C6H14 reacts with 19 mol of O2

for 0.1625 mol of C6H14, 1.5434 mol of O2 is required

But we have 0.4875 mol of O2

so, O2 is limiting reagent

we will use O2 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 = (12/19)* moles of O2

= (12/19)*0.4875

= 0.3079 mol

mass of CO2 = number of mol * molar mass

= 0.3079*44.01

= 13.6 g

Answer: 13.6 g