Final answer:
The mole fraction of CO2 in the products of complete butane combustion with excess air is the number of moles of CO2 divided by the total moles of gas present. The calculation involves the stoichiometry of the butane combustion reaction and considering the composition of air, but a numerical value cannot be provided without the specific 'x' moles of air supplied.
Step-by-step explanation:
When butane (C4H10) is burned completely with excess air, the reaction produces carbon dioxide (CO2) and water vapor (H2O). The balanced chemical equation for this combustion is:
C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
To calculate the mole fraction of CO2 in the reaction products, we first need to know the stoichiometry from the balanced equation. For every mole of butane combusted, 4 moles of CO2 are produced. If we're assuming that 'x' moles of air (which contains oxygen) are supplied for every mole of butane, and considering that air is about 21% oxygen by volume, we can then determine the amount of oxygen used in the reaction. However, as the question provides that there's excess air, we'll focus on the CO2 produced from the butane.
The mole fraction of CO2 is calculated by dividing the number of moles of CO2 by the total moles of gas in the products. Because the reaction is complete combustion with excess air, the only other significant component in the products is nitrogen (N2) from the air, which has not reacted. Assuming air is roughly 79% nitrogen and 21% oxygen, the majority of the air remains as N2 after the reaction.
Therefore, the mole fraction of CO2 is approximately:
4 moles CO2 / (4 moles CO2 + (4 * 3.76) moles N2) = 4 / (4 + (4 * 3.76))
As the exact mole fraction depends on the total amount of air supplied, which is not specified, we can't give a numerical value. However, the approach to calculate it is as described.