Question

Liquid octane (CH3(CH2)6CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 4.17 g of water is produced from the reaction of 9.14 g of octane and 23.4 g of oxygen gas, calculate the percent per yield of water. Round your answer to 3 significant figures.

Answer 1

The octane is oxidsed as per following balanced chemical equation

2CH₃(CH₂)₆CH₃ + 25 O₂ --------> 18H₂O (g) + 16CO₂(g)

moles of octane = mass of octane / molar mass = 9.14 / 114 = 0.08 mol

Moles of oxygen = 23.4g / 32g /mol = 0.73 mol

2 mol of octane need 25 mol of oxygen

0.08 mol of octane will need 0.08 X 25 = 2 mol of oxygen

But there are onlt 0.73 mol of oxygen. Hence oxygen is present in lesser amount than what is required as per balanced equation. Hence oxygen is the limiting reagent, it will determine the amount of product.

Theoretical yield : It is the amount of product that is expected to form the given amount of reactant.

25 mol or 800g of oxygen form 18 mol or 324g of water

1 g of oxygen will form 324 / 800 = 0.405 g of water

Therefore 23.4 g of oxygen will form 0.405 X 23.4g = 9.477g water

9.477 g water is the theoretical yield.

Percent yield = (amount of product formed / theoretical yield ) X 100

Percent yield = (4.17 / 9.477) X 100 = 44%