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.
