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How many moles of CO2 will be produced by the combustion of 3.00 moles of heptane C7H16 with 10.0 moles of O2?

User Mahes
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Final answer:

For the combustion of 3.00 moles of heptane with 10.0 moles of O2, the oxygen is the limiting reagent and only 6.36 moles of CO2 can be produced.

Step-by-step explanation:

Stoichiometry of Heptane Combustion

The combustion of heptane (C7H16) is a chemical reaction where heptane reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). To find out how many moles of CO2 would be produced from the combustion of 3.00 moles of heptane with 10.0 moles of O2, we need the balanced chemical equation:

C7H16 + 11O2 → 7CO2 + 8H2O

From the balanced equation, we see that each mole of heptane produces 7 moles of CO2. However, the combustion reaction requires 11 moles of oxygen to completely react with 1 mole of heptane. Given that there are only 10 moles of O2 available, the reaction is limited by the amount of oxygen. Therefore, we cannot use the full 3 moles of heptane, and instead we must calculate how many moles of heptane can react with the available oxygen:

10 moles O2 × (1 mole C7H16 / 11 moles O2) = 0.909 moles C7H16

Using the stoichiometric coefficients from the balanced equation, we can find the moles of CO2 produced from 0.909 moles of heptane:

0.909 moles C7H16 × (7 moles CO2 / 1 mole C7H16) = 6.36 moles CO2

Therefore, with 10.0 moles of O2, 6.36 moles of CO2 will be produced from the combustion of heptane. This is the maximum amount of CO2 that can be produced given the constraints of the reaction.

User Doron Cohen
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