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Liquid octane (CH3(CH2)6CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 36.3g of water is produced from the reaction of 30.84g of octane and 137.2g of oxygen gas, calculate the percent yield of water.

User Onestop
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Answer:Write the balanced chemical equation for the reaction:

2 C8H18 (octane) + 25 O2 → 16 CO2 + 18 H2O

Calculate the molar masses of the substances involved:

Molar mass of octane (C8H18) = 8 * (12.01 g/mol) + 18 * (1.01 g/mol) ≈ 114.23 g/mol

Molar mass of oxygen gas (O2) = 2 * (16.00 g/mol) ≈ 32.00 g/mol

Molar mass of water (H2O) = 2 * (1.01 g/mol) + 16.00 g/mol ≈ 18.02 g/mol

Determine the limiting reactant:

To find the limiting reactant, we need to compare the moles of octane and oxygen gas. Let's calculate the number of moles of each:

Moles of octane = Mass of octane / Molar mass of octane

Moles of octane = 30.84 g / 114.23 g/mol ≈ 0.2701 mol

Moles of oxygen gas = Mass of oxygen gas / Molar mass of oxygen gas

Moles of oxygen gas = 137.2 g / 32.00 g/mol ≈ 4.2875 mol

According to the balanced equation, the stoichiometric ratio between octane and oxygen gas is 2:25. Since the stoichiometric ratio is larger for oxygen gas (25) compared to octane (2), oxygen gas is the limiting reactant.

Calculate the theoretical yield of water:

Using the stoichiometric ratio, we can calculate the theoretical yield of water from the moles of oxygen gas:

Moles of water produced = Moles of oxygen gas * (18 moles of water / 25 moles of oxygen gas)

Moles of water produced = 4.2875 mol * (18/25) ≈ 3.096 mol

Calculate the theoretical mass of water:

Theoretical mass of water = Moles of water produced * Molar mass of water

Theoretical mass of water = 3.096 mol * 18.02 g/mol ≈ 55.777 g

Calculate the percent yield:

Percent yield = (Actual yield / Theoretical yield) * 100

Percent yield = (36.3 g / 55.777 g) * 100 ≈ 65.01%

The percent yield of water in this reaction is approximately 65.01%.

Step-by-step explanation:

User Dankyi Anno Kwaku
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