Answer to Question #266529 in Chemistry for Roseanne

Question

Liquid octane CH

3

CH

2

6

CH

3

 will react with gaseous oxygen O

2

 to produce gaseous carbon dioxide CO

2

 and gaseous water H

2

O

. Suppose 1.14 g of octane is mixed with 7.1 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

Solution:

The molar mass of octane is 114 g/mol

The molar mass of O₂ is 32 g/mol

 

Determine moles of each reactant:

(1.14 g octane) × (1 mol octane / 114 g octane) = 0.010 mol octane

(7.1 g O₂) × (1 mol O₂ / 32.00 g O₂) = 0.222 mol O₂

Balanced chemical equation:

2CH₃(CH₂)₆CH₃(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)

According to the chemical equation above:

2 mol of octane reacts with 25 mol of O₂

Thus, 0.010 mol of octane reacts with:

(0.010 mol octane) × (25 mol O₂ / 2 mol octane) = 0.125 mol O₂

However, initially there is 0.222 mol of O₂ (according to the task).

Therefore, octane acts as limiting reagent and O₂ is excess reagent.

 

According to stoichiometry:

2 mol of octane produces 16 mol of CO₂

Thus, 0.010 mol of octane produces:

(0.010 mol octane) × (16 mol CO₂ / 2 mol octane) = 0.08 mol CO₂

The molar mass of CO₂ is 44 g/mol

Therefore,

(0.08 mol CO₂) × (44 g CO₂ / 1 mol CO₂) = 3.52 g CO₂

Answer: The mass of carbon dioxide (CO₂) is 3.52 grams