Question

Liquid hexane CH3CH24CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 0.664g of water is produced from the reaction of 1.7g of hexane and 3.2g of oxygen gas, calculate the percent yield of water. Round your answer to 2 significant figures.

Answer 1

The chemical reaction between Liquid hexane and gaseous oxygen is-

2 CH₃(CH₂)₄CH₃ (l) + 19O₂(g) ------------> 12CO₂(g) + 14H₂O(g)

i.e 2 moles of Hexane will react with 19 moles of oxygen to produce 12 moles of gaseous carbon dioxide and 14 moles water

Now given

mass of hexane taken = 1.7g

So mols of hexane taken = mass / molar mass of hexane

= 1.7g / 86.18 g/mol

= 0.0197 mols

That means mols of Oxygen required fro complete reaction = 19/2 * 0.0197

= 0.187 mols

Now given mass of oxygen taken = 3.2g

So mols of oxygen taken = mass / molar mass of oxygen

= 3.2g / 32 g/mol

= 0.1 mols

That means we have less than required mols of oxygen gas present. So here the limiting reagent will be Oxygen and the complete reaction will be as per the mols of Oxygen involved. i.e

mols of Oxygen reacted = 0.1 mols

mols of hexane reacted = 2/19 * 0.1 = 0.0105 mols

mols of H2O to be formed = 14/19 * 0.1 = 0.074 mols

That means mass of  H2O to be formed = moles * molar mass of H2O

= 0.074 mols * 18 g/mole

= 1.332 g

But given actual yeild of water = 0.664g

So % yeild = (theoritical yeild - actua yeild) / theoritical yeild * 100

= (1.332 g - 0.664g) / 1.332 g * 100

= 50.15 %

= 50 %