How many liters of c7h16 would be required to form 20.0 g of h2o

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