Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 17.2 g of hexane is mixed with 19. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

16.5g of CO₂ could be produced

Explanation:

The combustion of hexane occurs as follows:

C₆H₁₄(l) + 19/2O₂ → 6CO₂ + 7H₂O

Where 1 mole of hexane reacts with 19/2 moles of O₂.

To solve this question we need to find the moles of each reactant in order to find limiting reactant. The moles of the limiting reactant will determine the moles of CO₂ produced:

Moles C₆H₁₄ -Molar mass: 86.18g/mol-:

17.2g hexane * (1mol / 86.18g) = 0.200 moles hexane

Moles O₂ -Molar mass: 32g/mol-:

19g O₂ * (1mol / 32g) = 0.594 moles oxygen.

For a complete reaction of 0.594 moles of oxygen are required:

0.594 moles O₂ * (1mol C₆H₁₄ / 19/2 moles O₂) = 0.0625 moles C₆H₁₄.

As there are 0.200 moles of hexane, hexane is the excess reactant and oxygen the limiting reactant.

The moles of CO₂ produced assuming a yield of 100% -All moles of oxygen react producing carbon dioxide.:

0.594 moles O₂ * (6mol CO₂ / 19/2 moles O₂) = 0.375 moles of CO₂ could be produced. The mass is:

0.375 moles of CO₂ * (44.01g / mol) =