Question

Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.17 g of hexane is mixed with 5.1 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

Answer 1

The most difficult part of these limiting reagent questions is to balance them, first balance the Carbons then Oxygen then hydrogen, in this order until all are balanced, you will have to balance several times:

2 C₆H₁₄ + 19 O₂ ----> 12 CO₂ + 14 H₂O

From mole concept, you have studied that 2 moles of hexane react with 19moles of oxygen to give 14 moles of water.

Calculate the Molar mass to find the number of moles by using the formula (no. of moles=given mass/molar mass)

Hexane = 86 g/mol, Oxygen = 32g/mol

Moles of hexane = 5.17 g/86 g/mol = 0.06 mol

Moles of oxygen = 5.1g/32 g/mol = 0.16 mol

Multiply the stoichiometry to the number of moles to see which one will be limiting. A limiting reagent is a reagent which ends first as the reaction goes on. this stops the reaction.

2 mol of hexane to react with 19 mol of water,

0.06 mol hexane if reacts completely, it will need 19*0.06/2=0.57 mol oxygen will be needed. But the oxygen present is only 0.16 mol.

Therefore we can see that oxygen is the limiting reagent.

[If while checking we find that more oxygen is present then we would have considered that hexane is limiting]

Using the unitary method:

19 mol oxygen gives 14 moles steam (water)

therefore 1mol oxygen gives 14/19 =0.74 moles water

0.16 mol oxygen gives 0.74mol* 0.16 mol = 0.12 moles of water.

Mass of water produced= moles*molar mass = 0.12mol * 18g/mol = 2.16 g steam