A mass of 8.15 g C2H4(g) reacts with excess oxygen. If 16.2 g CO2(g) is collected, what is the percent yield of the reaction? C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)Group of answer choices41.0%31.7%57.1%63.3%
Question
A mass of 8.15 g C2H4(g) reacts with excess oxygen. If 16.2 g CO2(g) is collected, what is the percent yield of the reaction? C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)Group of answer choices41.0%31.7%57.1%63.3%
Solution
To solve this problem, we need to follow these steps:
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Calculate the molar mass of C2H4 and CO2. The molar mass of C2H4 is approximately 28.05 g/mol and the molar mass of CO2 is approximately 44.01 g/mol.
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Calculate the number of moles of C2H4 and CO2. The number of moles is calculated by dividing the mass by the molar mass. So, for C2H4, it's 8.15 g / 28.05 g/mol = 0.290 moles. For CO2, it's 16.2 g / 44.01 g/mol = 0.368 moles.
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According to the balanced chemical equation, 1 mole of C2H4 produces 2 moles of CO2. So, 0.290 moles of C2H4 should theoretically produce 0.290 * 2 = 0.580 moles of CO2.
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The percent yield is calculated by dividing the actual yield (the number of moles of CO2 collected) by the theoretical yield (the number of moles of CO2 that should have been produced) and then multiplying by 100. So, the percent yield is (0.368 moles / 0.580 moles) * 100 = 63.3%.
So, the percent yield of the reaction is 63.3%.
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