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jekas [21]
3 years ago
8

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

Chemistry
1 answer:
professor190 [17]3 years ago
5 0

Answer:

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: <em>C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.</em>

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: <em>n = mass/molar mass.</em>

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

<em><u>Using cross multiplication:</u></em>

1.0 mol of heptane releases → 8 moles of water.

0.7 mol of heptane releases → ??? moles of water.

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

<em>∴ When 69.9 g heptane is burned it releases </em><em>5.6</em><em> mol water. </em>

<em />

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