Fe can be prepared as 2 Al + Fe2O3 ----> 2 Fe + Al2O3. Suppose that 0.450 moles of Fe2O3 are reacted with an excess of Al. Suppose that 43.6 grams of Fe are obtained. What is the percent yield of Fe?
a) 16.2%
b) 60.7%
c) 86.7%
d) 43.4%
Percent yield= actual yield of product divided by theoretical yield of product times 100.
The actual yield is stated to be 43.6 g of Fe, so I just need to know how to get the theoretical yield from the information it already gives me from the problem.
theoritical yield: .45*(2/1) moles, so you get .90 moles Fe or .90*55 =about 50 grams in my head.
percent yield=44/50= about 88 percnet. Answer c.
86.7%
Well, it looks like this problem is all about chemistry, but let's add a touch of Clown Bot humor to make it more interesting!
To find the theoretical yield, we need to first convert the 0.450 moles of Fe2O3 into moles of Fe. According to the balanced equation, the ratio is 1:2. So, we have:
0.450 moles Fe2O3 * (2 moles Fe / 1 mole Fe2O3) = 0.900 moles Fe
Now it's time to convert moles of Fe into grams. We need to find the molar mass of Fe, which is 55.845 g/mol. So:
0.900 moles Fe * (55.845 g Fe / 1 mole Fe) = 50.2615 g Fe (approximately)
And voila! The theoretical yield is approximately 50.2615 grams of Fe. Now we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100
= (43.6 g Fe / 50.2615 g Fe) * 100
≈ 86.7%
So, the correct answer is (c) 86.7%. Just remember, my jokes have a higher yield than my chemistry knowledge!
To find the theoretical yield of Fe, we need to calculate how many moles of Fe can be obtained from the reaction.
The balanced equation tells us that 2 moles of Al react with 1 mole of Fe2O3 to produce 2 moles of Fe. This means that the stoichiometry of the reaction is 2:1 for Fe:Fe2O3.
Given that we have 0.450 moles of Fe2O3, we can use this stoichiometry to calculate the theoretical yield of Fe.
1 mole of Fe2O3 produces 2 moles of Fe.
Therefore, 0.450 moles of Fe2O3 will produce 2 x 0.450 moles of Fe.
Theoretical yield of Fe = 2 x 0.450 moles = 0.900 moles
Now that we have the theoretical yield in moles, we can convert it to grams using the molar mass of Fe.
The molar mass of Fe is approximately 55.85 g/mol.
Theoretical yield of Fe = 0.900 moles x 55.85 g/mol = 50.26 grams
Therefore, the theoretical yield of Fe is 50.26 grams.
To calculate the percent yield, we can use the formula:
Percent yield = (actual yield / theoretical yield) x 100
Given that the actual yield is 43.6 grams and the theoretical yield is 50.26 grams:
Percent yield = (43.6 / 50.26) x 100 ≈ 86.7%
Therefore, the percent yield of Fe is approximately 86.7%.
The correct answer is option c) 86.7%.