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Show below is the combustion of hexane, C6H14.C6H14 + O2--> CO2 + H2OIf 420 grams of hexane and 2,200 grams of air are supplied, compute for the percentage excess.Write your final answer in two decimal places.Use the following mass numbers:C - 12 g/molH - 1 g/molO - 16 g/mol

Question

Show below is the combustion of hexane, C6H14.C6H14 + O2--> CO2 + H2OIf 420 grams of hexane and 2,200 grams of air are supplied, compute for the percentage excess.Write your final answer in two decimal places.Use the following mass numbers:C - 12 g/molH - 1 g/molO - 16 g/mol

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Solution

To solve this problem, we need to follow these steps:

  1. First, we need to balance the chemical equation for the combustion of hexane. The balanced equation is: 2C6H14 + 19O2 --> 12CO2 + 14H2O

  2. Next, we calculate the molar mass of hexane (C6H14). This is (612) + (141) = 86 g/mol.

  3. We then find out how many moles of hexane are present in 420 grams by dividing the mass of hexane by its molar mass. This gives us 420/86 = 4.88 moles of hexane.

  4. According to the balanced equation, 1 mole of hexane requires 19/2 = 9.5 moles of O2 for complete combustion. So, 4.88 moles of hexane would require 4.88 * 9.5 = 46.36 moles of O2.

  5. The molar mass of O2 is (2*16) = 32 g/mol. So, 46.36 moles of O2 would weigh 46.36 * 32 = 1483.52 grams.

  6. The air is 21% oxygen by mass. So, in 2200 grams of air, the mass of oxygen is 0.21 * 2200 = 462 grams.

  7. The number of moles of O2 in 462 grams is 462/32 = 14.44 moles.

  8. The percentage excess of O2 is ((14.44 - 46.36) / 46.36) * 100 = -68.85%.

However, since the percentage cannot be negative, we conclude that there is no excess O2. Instead, there is a deficiency of O2.

This problem has been solved

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