Answer:
52.1 g is the maximum mass of CO₂, that can be produced by this combustion
Explanation:
Mass of Octane: 63 g
Mass of O₂: 59.4 g
This is a combustion reaction where the products, are always water and CO₂. We define the equation:
2C₈H₁₈ (l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)
As we have, both mases of each reactant, we must define which is the limiting reagent. We convert the mass to moles:
63 g. 1mol / 114g = 0.552 moles
59.4 g . 1mol / 32g = 1.85 moles
Certainly, the limiting reagent is the oxygen:
2 moles of octane need 25 moles of O₂ to react
Therefore, 0.552 moles of octane must need (0.552 . 25) /2 = 6.9 moles of O₂ (I do not have enough moles of oxygen, I need 6.9 and I only got 1.85 moles)
When we know the limiting reagent we can do the calculations with the stoichiometry of the reaction:
25 moles of O₂ can produce 16 moles of CO₂
Therefore, 1.85 moles of O₂ may produce (1.85 . 16) /25 = 1.18 moles.
We convert the moles to mass to get the final answer:
1.18 mol . 44 g / 1mol = 52.1 g
