Question

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose 73. g of octane is mixed with 105. 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)= 73.0 g

number of mol of C8H18,

n = mass of C8H18/molar mass of C8H18

=(73.0 g)/(114.224 g/mol)

= 0.6391 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 105.0 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(105.0 g)/(32 g/mol)

= 3.281 mol

Balanced chemical equation is:

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

2 mol of C8H18 reacts with 25 mol of O2

for 0.639095 mol of C8H18, 7.988689 mol of O2 is required

But we have 3.28125 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

According to balanced equation

mol of C8H18 reacted = (2/25)* moles of O2

= (2/25)*3.28125

= 0.2625 mol

mol of C8H18 remaining = mol initially present - mol reacted

mol of C8H18 remaining = 0.639095 - 0.2625

mol of C8H18 remaining = 0.376595 mol

mass of C8H18,

m = number of mol * molar mass

= 0.3766 mol * 114.224 g/mol

= 43 g

Answer: 43 g