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Answer :
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%.
Explanation:
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