Answer:
36.9 g is the mass of octane that is left over
Explanation:
We state the reaction by the information given.
Octane as a reactant → C₈H₁₈
Oxygen as the other reactant → O₂
We have both masses, so we can predict the limiting reactant
We convert the mass to moles:
94 g / 114g/mol = 0.824 moles of octane
200 g / 32 g/mol = 6.25 moles of O₂
The reaction is:
2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)
2 moles of octane need 25 moles of O₂ to react
Then, 0.824 moles of octane may react with (0.824 . 25) /2 = 10.3 moles O₂
We have 6.25 moles and we need 10.3, then the O₂ is the limiting reagent
Octane is the excess reactant:
25 moles of O₂ may react with 2 moles of octane
Therefore, 6.25 moles of O₂ will react with (6.25 . 2) / 25 = 0.500 moles
We have 0.824 moles of octane and we only need 0.500, therefore
(0.824-0.500) = 0.324 moles are left over by the reaction.
We convert the moles to mass → 0.324 mol . 114g / 1 mol = 36.9 g
