Given:C2H6(g) + O2(g) ⟶ CO2(g) + H2O(g)If 395 grams of ethane and 1,595 grams of air are supplied, compute for the % excess.Write your final answer in two decimal places.Use the following mass numbers:O - 16C - 12H - 1
Question
Given:C2H6(g) + O2(g) ⟶ CO2(g) + H2O(g)If 395 grams of ethane and 1,595 grams of air are supplied, compute for the % excess.Write your final answer in two decimal places.Use the following mass numbers:O - 16C - 12H - 1
Solution
To solve this problem, we need to follow these steps:
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First, we need to balance the chemical equation. The balanced equation is: 2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g)
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Next, we calculate the molar mass of each reactant. For C2H6, it's (212) + (61) = 30 g/mol. For O2, it's 2*16 = 32 g/mol.
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Then, we calculate the number of moles for each reactant. For C2H6, it's 395 g / 30 g/mol = 13.17 mol. For O2, it's 1595 g / 32 g/mol = 49.84 mol.
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According to the balanced equation, the mole ratio of C2H6 to O2 is 2:7. So, for every 2 moles of C2H6, we need 7 moles of O2. Therefore, for 13.17 moles of C2H6, we need 13.17 * (7/2) = 46.09 moles of O2.
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Now, we compare the required amount of O2 with the supplied amount. We have 49.84 moles of O2 supplied, but we only need 46.09 moles. So, the excess is 49.84 - 46.09 = 3.75 moles.
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Finally, we calculate the percentage of excess O2. It's (3.75 / 49.84) * 100 = 7.52%.
So, the percentage of excess O2 is 7.52%.
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