0.0215 L
The combustion of C₇H₁₆ is given by the balanced equation;
C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O
From the question we are given;
Mass of H₂O that would be formed as 20.0 g
We are required to determine the number of liters of C₇H₁₆ that would produce 20.0 g of water.
We are going to use the following steps;
Number of moles = Mass ÷ Molar mass
Molar mass of H₂O = 18.02 g/mol
= 20.0 g ÷ 18.02 g/mol
= 1.1099 moles
From the equation, 1 mole of C₇H₁₆ produces 8 moles of H₂O
Therefore, the mole ratio of C₇H₁₆ to H₂O is 1 : 8
Thus, the moles of C₇H₁₆ = Moles of water ÷ 8
= 1.1099 moles ÷ 8
= 0.139 moles
Mass = Number of moles × Molar mass
Molar mass of C₇H₁₆ =100.21 g/mol
Therefore;
Mass = 0.139 moles × 100.21 g/mol
= 13.929 g
Density of Heptane is 648 Kg/m³ or 0.648 g/mL
Therefore; since Volume = Mass ÷ Density
Then; Volume of heptane = 13.929 g ÷ 0.648 g/mL
= 21.495 mL
but, 1000 mL = 1 L
Therefore, the volume of heptane = 0.021495 L
= 0.0215 L
The volume of Heptane required will be 0.0215 L
