Question

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose 2.3 g of octane is mixed with 14.2 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Molar mass of C8H18,

MM = 8*MM(C) + 18*MM(H)

= 8*12.01 + 18*1.008

= 114.224 g/mol

mass(C8H18)= 2.3 g

number of mol of C8H18,

n = mass of C8H18/molar mass of C8H18

=(2.3 g)/(114.224 g/mol)

= 2.014*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:

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

2 mol of C8H18 reacts with 25 mol of O2

for 0.0201 mol of C8H18, 0.2517 mol of O2 is required

But we have 0.4437 mol of O2

so, C8H18 is limiting reagent

So, there won’t be any octane left over

Answer: 0 g