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:

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

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.

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.

it is the most commonly accepted one