Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 2.3 g of octane is mixed with 12.4 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Round your answer to 2 significant digits.

Octane will be used completely.

Explanation:

To calculate the moles :

The balanced chemical reaction will be

According to stoichiometry :

2 moles of octane require = 25 moles of

Thus 0.0202 moles of octane will require=  of

Thus octane is the limiting reagent as it limits the formation of product and is the excess reagent.  

Thus octane will be used completely.

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