Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.7 g of octane is mixed with 27.0 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

The minimum mas of octane that could be left over by the chemical reaction is nothing because every mole of octane will react with oxygen.

Explanation:

The reaction of octane (C₈H₁₈: Molar mass 12,01×8+1,01×18 = 114,26g/mol) with oxygen (O₂: Molar mass 16×2 = 32g/mol) produce:

C₈H₁₈ + O₂ → CO₂ + H₂O

The balanced reaction is:

C₈H₁₈ + 25/2 O₂ → 8CO₂ + 9H₂O

5,7g of octane are:

5,7g × (1mol / 114,26g) = 0,050 moles of octane

And 27,0g of Oxygen are:

27,0g × (1mol / 32g) = 0,844 moles of oxygen.

Based on the reaction, the theoretical moles of octane that react with 0,844 moles of oxygen are:

0,844 moles of O₂ × ( 1 mole octane / 25/2 mol O₂) = 0,068 moles of octane

As moles of octane you have are 0,050 moles, the minimum mas of octane that could be left over by the chemical reaction is nothing because every mole of octane will react with oxygen.

I hope it helps!

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