Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 8.00 g of octane is mixed with 37. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

11.3 g of H₂O will be produced.

Explanation:

The combustion is:

2C₈H₁₈ + 25O₂→ 16CO₂ + 18H₂O

First of all, we determine the moles of the reactants in order to find out the limiting reactant.

8 g / 114g/mol = 0.0701 moles of octane

37g / 32 g/mol = 1.15 moles of oxygen

The limiting reagent is the octane. Let's see it by this rule of three:

25 moles of oxygen react to 2 moles of octane so

1.15 moles of oxygen will react to (1.15. 2) / 25 = 0.092 moles of octane.

We do not have enough octane, we need 0.092 moles and we have 0.0701 moles. Now we work with the stoichiometry of the reaction so we make this rule of three:

2 moles of octane produce 18 moles of water

Then 0.0701 moles of octane may produce (0.0701. 18) / 2 = 0.631 moles of water.

We convert the moles to mass → 0.631 mol. 18 g/1mol = 11.3 g of H₂O will be produced.