Write a balanced chemical equation, including physical state symbols, for the combustion of liquid heptane into gaseous carbon dioxide and gaseous water. C7H6) + 110, (g) 7 16 7CO2 (g) 8H20 (8) 2. Suppose 0.140 kg of heptane are burned in air at a pressure of exactly 1 atm and a temperature of 11.0 °C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.

The volume of carbon dioxide gas that is produced is 243L

Explanation:

This is the equation:

2C₇H₆ (g) + 17O₂ (g) → 14CO₂ (g) + 6H₂O (g)

We have the mass of heptane that was burned, so let's determine the moles

Mass (g) / molar mass

We convert the mass to g → 0.140 kg. 1g / 1*10⁻³ kg = 140 g

140 g / 90 g/mol = 1.55 moles

Ratio is 2:14. 2 moles of heptane can produce 14 moles of CO₂

Therefore 1.55 moles, would produce (1.55. 14) / 2 = 10.9 moles of CO₂

Finally we have to work with density if we want to know volume. Firstly we convert the moles of CO₂ to g.

10.9 moles of CO₂. 44 g/mol = 479.6 g of CO₂

CO₂ density = CO₂ mass / CO₂ volume

CO₂ volume = CO₂ mass / CO₂ density

CO₂ volume = 479.6 g / 0,001976 g/mL → 242712.5 mL

Let's convert the volume to L

242712.5 mL. 1L/1000 mL = 242.71 L

Rounded to 3 significant digits → 243L