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.

because natural gas is colorless, odorless, and tasteless, distributors add mercaptan (a chemical that smells like sulfur) to give natural gas a distinct unpleasant odor (it smells like rotten eggs). they locate the types of rock that are likely to contain natural gas deposits.

it would be a gas formation