Question

How many grams of CO2 are produced by the combustion of 289 g of a mixture that is 28.6% CH4 and 71.4% C3H8 by mass?

Answer 1

Answer:

• 433g

Explanation:

1. Calculate the mass of each reactant gas in the mixture

a) CH₄

• 28.6% of 289g
• 0.286 × 289g = 82.654g

b) C₃H₈

• 71.4% of 289g
• 0.714 289g = 206.346g

2. Calculate the mass of CO₂ produced by each gas in the mixture

a) CH₄

i) Balanced chemical equation:

• CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

ii) Number of moles of CH₄:

• number of moles = mass in grams / molar mass
• molar mass of CH₄ = 16.04g/mol
• number of moles = 82.654g / 16.04 g/mol = 5.153 mol

iii) Number of moles of CO₂

• From the balanced chemical equation, the mole ratio is 1 mol CH₄ : 1 mol CO₂.
• Hence, 5.153 mol of CO₂ are produced

iv) Convert the number of moles to mass

• mass = number of moles × molar mass
• molar mass of CO₂ = 44.01g/mol
• mass = 5.153 mol × 44.01g/mol = 226.8 g

b) C₃H₈

i) Balanced chemical equation:

• C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

ii) Number of moles of C₃H₈:

• number of moles = mass in grams / molar mass
• molar mass of C₃H₈ = 44.1g/mol
• number of moles = 206.346g / 44.1 g/mol = 4.679 mol

iii) Number of moles of CO₂

• From the balanced chemical equation, the mole ratio is 1 mol C₃H₈ : 1 mol CO₂.
• Hence, 4.679 mol of CO₂ are produced

iv) Convert the number of moles to mass

• mass = number of moles × molar mass
• molar mass of CO₂ = 44.01g/mol
• mass = 4.679 mol × 44.01g/mol = 205.9 g

3. Total mass of CO₂

Add the two values found above:

• 226.8g + 205.9g = 432.7g

Round to 3 significant figures: 433g.