Answer:
Maximum amount of carbon dioxide that can be formed → 7.52 g
Limiting reactant  → O₂
Amount of the excess reagent, after the reaction occurs → 12.9 g
Explanation:
We determine the reaction. This is a combustion:
2C₆H₆ (l) + 15O₂ (g) → 6CO₂(g) + 6H₂O (g)
We need to determine the limting reactant so we convert the mass to moles:
8.70 g. 1mol / 78g = 0.111 moles of benzene
13.7 g . 1mol / 32g = 0.428 moles of oxygen
Ratio is 2:15. 2 moles of benzene react with 15 moles of Oâ‚‚
Then, 0.111 moles of benzene may react with (0.111 .15) /2 = 0.832 moles of Oâ‚‚
We have 0.428 moles but we need 0.832 moles for the complete reaction, so there are (0.832 - 0.428) = 0.404 moles remaining. Oxygen is the limiting reactant. We work now, with the reaction:
15 moles of Oâ‚‚ can produce 6 moles of COâ‚‚
So, 0.428 moles of Oâ‚‚ may produce (0.428 . 6)/ 15 = 0.171 moles of COâ‚‚
We convert the moles to mass → 0.171 mol . 44g / 1mol = 7.52 g
This is the maximum amount of carbon dioxide that can be formed
We convert the mass of the limiting reactant that remains after the reaction is complete → 0.404 mol . 32g / 1mol = 12.9 g of O₂
