Answer:
9 liters of CO₂ are produced by this combustion
Explanation:
In order to determine the volume of produced CO₂, we start with the reaction:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)
We need, O₂ density to find out the mass, that has reacted.
δ O₂ = O₂ mass / O₂ volume → δ O₂ . O₂ volume = O₂ mass
δ O₂ = 1.429 g /dm₃ (1dm³ = 1L) 1.429 g/L . 15L = 21.4 g of O₂
We convert the mass to moles: 21.4 g . 1mol / 32 g = 0.670 moles
By stoichiometry, 5 moles of O₂ can produce 3 moles of CO₂
Then, 0.670 moles of O₂ will produce (0.670 . 3) /5 = 0.402 moles of dioxide.
We apply Ideal Gases Law for STP, to find out the CO₂ volume
V = (n . R . T) / P → V = (0.402 mol . 0.082 . 273K) / 1 atm = 8.99 L ≅ 9 L
