The amount of carbon dioxide (CO2) produced by the combustion of a specific amount of heptane depends on the stoichiometry of the reaction. Heptane has a molecular formula of C7H16.
The balanced chemical equation for the combustion of heptane can be written as:
C7H16 + 11O2 → 7CO2 + 8H2O
From the equation, we can see that 1 mole of heptane (C7H16) reacts with 11 moles of oxygen (O2) to produce 7 moles of carbon dioxide (CO2) and 8 moles of water (H2O).
To calculate the amount of CO2 produced, we need to know the molar volume of an ideal gas at standard temperature and pressure (STP), which is approximately 22.4 liters per mole.
First, we need to determine the number of moles of heptane in 10 liters. Since heptane has a molar volume of approximately 100.21 grams per mole, we can calculate the number of moles using the following equation:
moles of heptane = volume (in liters) / molar volume (in liters per mole)
moles of heptane = 10 L / 100.21 L/mol ≈ 0.0998 moles
According to the balanced equation, 1 mole of heptane produces 7 moles of carbon dioxide. Therefore, the number of moles of carbon dioxide produced is:
moles of CO2 = moles of heptane × 7 = 0.0998 moles × 7 ≈ 0.6986 moles
Finally, we can convert the moles of CO2 to volume (in liters) using the molar volume of an ideal gas at STP:
volume of CO2 = moles of CO2 × molar volume (in liters per mole)
volume of CO2 = 0.6986 moles × 22.4 L/mol ≈ 15.65 liters
Therefore, the combustion of 10 liters of heptane produces approximately 15.65 liters of carbon dioxide.