How many liters of C7H16 would be required to form 20.0 g of h2o

0.0215 L

Explanation:

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;

Step 1: Calculate the number of moles of H₂O

Number of moles = Mass : Molar mass

Molar mass of H₂O = 18.02 g/mol

= 20.0 g : 18.02 g/mol

= 1.1099 moles

Step 2: Moles of C₇H₁₆

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

Step 3: Calculate the mass of C₇H₁₆

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

Step 4 : Volume of C₇H₁₆ in liters

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