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.

aquifers are bounded above and below by impermeable rock. the driving force of groundwater movement in confined aquifers is pressure, rather than gravity.