Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.48 g of hexane is mixed with 43. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

No mass of hexane could be left over by the chemical reaction.

C₆H₁₄ is the limitin reagent

Explanation:

This is a combustion reaction were the oxygen is one of the reactant and it burns a compound in order to generate water and carbon dioxide.

In this case, we have liquid hexane combustion, so the reaction is:

2C₆H₁₄(l) +  19O₂(g)  →   12CO₂(g) +  14H₂O(g)

In this situation we are asked for the mass of a reactant that could be left over, this is the excess reagent.

We convert the masses of reactants to moles:

9.48 g . 1mol  / 86g = 0.110 moles of C₆H₁₄

43 g . 1 mol/32 g =  1.34 moles of O₂

The hexane may be the excess reagent but we confirm like this:

19 moles of O₂ need 2 moles of hexane to react

Then, 1.34 moles of O₂ will react with (1.34 . 2) / 19 = 0.141 moles

We need 0.141 moles, and we only have 0.110. Hexane is the limiting reagent so no mass could be left over by the chemical reaction.

In conclussion, oxygen is excess reactant. We verify:

2 moles of hexane need 19 moles of O₂ to react

Then, 0.110 moles of hexane will react with (0.110 . 19) / 2 = 1.04 moles of O₂

As I have 1.34 moles of oxygen, value is higher so in this case we are having mass that could be left over by the reaction.