Question

16 grams of propane, C 3 H 8 and 20 grams of oxygen, O 2 are reacted to produce carbon dioxide and water. Calculate the volume of CO 2 produced?

Answer 1

Answer:

The volume of CO₂ produced is 8.4 L

Explanation:

The mass of propane in the reaction C₃H₈ = 20 grams

The mass of, oxygen, O₂ in the reaction = 20 grams

The produce of the reaction are carbon dioxide, CO₂ and water, H₂O

The balanced equation of the (combustion) reaction can be presented as follows;

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

Therefore, one mole of propane, C₃H₈, reacts with five moles of oxygen, O₂, to produce three moles of carbon dioxide, CO₂, and four moles of water molecules, H₂O, as steam

The number of moles = Mass/(Molar mass)

The molar mass of propane, C₃H₈ = 44.1 g/mol

The number of moles of propane in 16 grams of propane = 16/44.1 ≈ 0.3628 moles

The molar mass of oxygen, O₂ = 32.0 g/mol

The number of moles of oxygen in 20 grams of propane = 20/32 ≈ 0.625 moles

Therefore;

Given that 1 mole of C₃H₈ reacts with 5 moles of O₂

1 mole of O₂ will react with 1/5 moles of C₃H₈

0.625 moles of O₂ will react with 0.625/5 = 0.125 moles of C₃H₈ to produce 3 × 0.125 = 0.375 moles of CO₂

1 mole of an ideal gas occupies 22.4 L at standard temperature and pressure

Taking CO₂ as an ideal gas, we have;

0.375 mole of CO₂ will occupy 0.375 × 22.4 L = 8.4 L

Therefore, the volume of CO₂ produced = The volume occupied by the 0.375 moles of CO₂ = 8.4 L.