When 69.9 g heptane is burned it releases __ mol water.

__ C7H16 + __ O2 → __ CO2 + __ H2O

1) When 69.9 g heptane is burned it releases 5.6 mol water.

2) C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.

Explanation:

Firstly, we should balance the equation of heptane combustion. The balanced equation is: C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.

This means that every 1.0 mole of complete combustion of heptane will release 8 moles of H₂O.

We need to calculate the no. of moles in 69.9 g of heptane that is burned using the relation: n = mass/molar mass.

n of 69.9 g of heptane = mass/molar mass = (69.9 g) / (100.21 g/mol) = 0.697 mol ≅ 0.7 mol.

Using cross multiplication:

1.0 mol of heptane releases → 8 moles of water.

0.7 mol of heptane releases →? moles of water.

∴ The no. of moles of water that will be released from burning (69.9 g) of water = (0.7 mol) (8.0 mol) / (1.0 mol) = 5.6 mol.

∴ When 69.9 g heptane is burned it releases 5.6 mol water.