Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 64. g of hexane is mixed with 72.0 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.

45.7g of CO₂ is the maximum amount of carbon dioxide that could be produced.

Explanation:

Hexane, C₆H₁₄, reacts with O₂ to produce H₂O and CO₂ as follows:

C₆H₁₄ + 13/2O₂ → 7H₂O + 3CO₂

Where 1 mole of hexane reacts with 13/2 moles of O₂.

To solve this question we need to convert the mass of each reactant to moles in order to find the limiting reactant. With limiting reactant we can find the maximum amount of CO₂ produced:

Moles C₆H₁₄ -Molar mass: 86.18g/mol-:

64g C₆H₁₄ * (1mol / 86.18g) = 0.7426 moles C₆H₁₄

Moles O₂ -Molar mass: 32g/mol-:

72.0g O₂ * (1mol / 32g) = 2.25 moles O₂

For a complete reaction of 2.25 moles of O₂ are required:

2.25 moles O₂ * (1mol C₆H₁₄ / 13/2 moles O₂) = 0.346 moles of C₆H₁₄

As there are 0.7426 moles of C₆H₁₄, C₆H₁₄ is the excess reactant and:

Oxygen is limiting reactant.

The maximum moles of CO₂ that could be produced are:

2.25 moles O₂ * (3mol CO₂ / 13/2 moles O₂) = 1.038 moles CO₂

In grams:

1.038 moles CO₂ * (44.01g / mol) =

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b