Show below is the combustion of hexane, C6H14.C6H14 + O2--> CO2 + H2OIf 407 grams of hexane and 2,239 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

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

1. Calculate the molar mass of hexane (C6H14). The molar mass of carbon (C) is 12 g/mol and hydrogen (H) is 1 g/mol. So, the molar mass of hexane is (612) + (141) = 86 g/mol.

2. Calculate the number of moles of hexane. The number of moles is given by the mass of the substance divided by its molar mass. So, the number of moles of hexane is 407 g / 86 g/mol = 4.73 moles.

3. The balanced chemical equation for the combustion of hexane is 2C6H14 + 19O2 --> 12CO2 + 14H2O. This means that 2 moles of hexane react with 19 moles of oxygen to produce 12 moles of carbon dioxide and 14 moles of water.

4. Calculate the required moles of oxygen for the reaction. From the balanced equation, we know that 2 moles of hexane require 19 moles of oxygen. So, 4.73 moles of hexane will require (4.73/2)*19 = 44.97 moles of oxygen.

5. Calculate the mass of the required oxygen. The molar mass of oxygen (O2) is 32 g/mol. So, the mass of the required oxygen is 44.97 moles * 32 g/mol = 1439.04 g.

6. The mass of air supplied is 2239 g. Air is approximately 21% oxygen by mass. So, the mass of oxygen in the air supplied is 21/100 * 2239 g = 470.19 g.

7. Calculate the percentage excess of oxygen. The percentage excess is given by ((actual amount - required amount) / required amount) * 100%. So, the percentage excess of oxygen is ((470.19 g - 1439.04 g) / 1439.04 g) * 100% = -67.32%.

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