Answer to Question #286140 in Chemistry for khryst

Question

4. A combustion reaction wherein 15.0 g of propane (C3H8) is burned with 75.0 g of oxygen (O2).

a. Determine the limiting reactant.

b. What is the maximum amount of CO2 (g) can you produce?.

Answer 1

Solution:

The molar mass of C₃H₈ is 44.1 g/mol

The molar mass of O₂ is 32.0 g/mol

Calculate the number of moles of each reactant:

(15.0 g C₃H₈) × (1 mol C₃H₈ / 44.1 g C₃H₈) = 0.34 mol C₃H₈

(75.0 g O₂) × (1 mol O₂ / 32.0 g O₂) = 2.34 mol O₂

Balanced chemical equation:

C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

According to stoichiometry:

1 mol of C₃H₈ reacts with 5 mol of O₂

Thus, 0.34 mol of C₃H₈ reacts with:

(0.34 mol C₃H₈) × (5 mol O₂ / 1 mol C₃H₈) = 1.7 mol O₂

However, initially there is 2.34 mol of O₂ (according to the task).

Thus, C₃H₈ acts as limiting reactant and O₂ is excess reactant.

According to stoichiometry:

1 mol of C₃H₈ produces 3 mol of CO₂

Thus, 0.34 mol of C₃H₈ produces:

(0.34 mol C₃H₈) × (3 mol CO₂ / 1 mol C₃H₈) = 1.02 mol CO₂

The molar mass of CO₂ is 44.01 g/mol

Hence,

(1.02 mol CO₂) × (44.01 g CO₂ / 1 mol CO₂) = 44.89 g CO₂

Answer:

a) The limiting reactant is propane (C₃H₈)

b) The maximum amount of CO₂ is 1.02 mol (or 44.89 g)