Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O Suppose 0.862 g of hexane is mixed with 5.1 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.

2.64g of CO₂ are produced

Explanation:

The reaction of hexane with O₂ is:

C₆H₁₄ + 19/2O₂ → 6CO₂ + 7H₂O

To solve this question, we must convert each mass of reactant to moles in order to find limiting reactant and, with the chemical reaction we can find the moles of carbon dioxide produced and the maximum mass produced:

Moles hexane -Molar mass: 86.18g/mol-:

0.862g * (1mol / 86.18g) = 0.01 moles

Moles Oxygen -Molar mass: 32g/mol-:

5.1g * (1mol / 32g) = 0.16 moles

For a complete reaction of 0.01 moles of hexane there are required:

0.01moles hexane * (19/2 Oxygen / 1mol Hexane) = 0.095 moles oxygen

As there are 0.16moles, limiting reactant is hexane.

0.01 moles of hexane produce:

0.01 moles * (6 moles carbon dioxide / 1 mole hexane) = 0.06 moles of CO₂.

The mass is -Molar mass CO₂: 44g/mol-:

0.06 moles of CO₂ * (44g/mol) =

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1 is the answer to this problem