Answer to Question #123506 in Chemistry for ayoub

Question

The complete combustion of octane with oxygen, represented by the unbalanced
chemical equation: C8H18 + O2 → CO2 + H2O
10 grams of octane and 10grams of oxygen will yield what mass of water?

Answer 1

Solution:

The balanced chemical equation:

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O 

According to the chemical equation: n(C₈H₁₈)/2 = n(O₂)/25 = n(H₂O)/18

Convert all given information into moles:

Moles of octane = n(C₈H₁₈) = Mass of C₈H₁₈ / Molar mass of C₈H₁₈

The molar mass of C₈H₁₈ is 114 g/mol

n(C₈H₁₈) = (10 g) / (114 g/mol) = 0.0877 mol

Moles of oxygen = n(O₂) = Mass of O₂ / Molar mass of O₂

The molar mass of O₂ is 32 g/mol

n(O₂) = (10 g) / (32 g/mol) = 0.3125 mol

Divide by coefficients of balanced equation:

octane ⇒ 0.0877 mol / 2 mol = 0.04385

oxygen ⇒ 0.3125 mol / 25 mol = 0.0125

Oxygen is the lower value. It is the limiting reactant.

We now use the limiting reactant to find the amount of water produced.

n(O₂)/25 = n(H₂O)/18

n(H₂O) = 18 × n(O₂) / 25

n(H₂O) = 18 × (0.3125 mol) / 25 = 0.225 mol

Moles of H₂O = n(H₂O) = Mass of H₂O / Molar mass of H₂O

The molar mass of H₂O is 18 g/mol.

Finally, the mass of H₂O is:

Mass of H₂O = (0.225 mol) × (18 g/mol) = 4.05 g

Answer: 4.05 grams of water will be produce.