Show below is the combustion of hexane, C6H14.C6H14 + O2--> CO2 + H2OIf 444 grams of hexane and 2,038 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). This is done by adding the molar masses of its constituent atoms, multiplied by their quantity in one molecule of hexane.

   Molar mass of C6H14 = (6 * 12 g/mol) + (14 * 1 g/mol) = 72 g/mol + 14 g/mol = 86 g/mol

2. Calculate the number of moles of hexane in 444 grams. This is done by dividing the mass of hexane by its molar mass.

   Moles of C6H14 = 444 g / 86 g/mol = 5.16 moles

3. The balanced chemical equation for the combustion of hexane is:

   2C6H14 + 19O2 --> 12CO2 + 14H2O

   From this equation, we can see that 2 moles of hexane react with 19 moles of oxygen. Therefore, 5.16 moles of hexane would react with:

   Moles of O2 = (19/2) * 5.16 = 49.02 moles

4. Calculate the mass of oxygen in 49.02 moles. This is done by multiplying the number of moles by the molar mass of oxygen.

   Mass of O2 = 49.02 moles * 32 g/mol = 1568.64 g

5. The mass of air supplied is 2038 grams. Air is approximately 21% oxygen by mass, so the mass of oxygen in the air supplied is:

   Mass of O2 in air = 0.21 * 2038 g = 428 g

6. The percentage excess of oxygen is then calculated by subtracting the mass of oxygen required for the reaction from the mass of oxygen in the air supplied, dividing by the mass of oxygen required for the reaction, and multiplying by 100.

   Percentage excess = ((428 g - 1568.64 g) / 1568.64 g) * 100 = -72.72%

However, since the percentage excess is negative, this means that there is actually a deficiency of oxygen. Therefore, there is no excess oxygen in this reaction.