For the following reaction, 8.70 grams of benzene (C6H6) are allowed to react with 13.7 grams of oxygen gas. benzene (C6H6) (l) + oxygen (g) carbon dioxide (g) + water (g) What is the maximum amount of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

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 O₂

Then, 0.111 moles of benzene may react with (0.111 .15) /2 = 0.832 moles of O₂

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 O₂ can produce 6 moles of CO₂

So, 0.428 moles of O₂ may produce (0.428 . 6)/ 15 = 0.171 moles of CO₂

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