when 4.50 g of fe2o3 is reduced with excess h2 in a furnace, 2.6 g of iron metal is recovered. What is the percent yield when molar mass of Fe2O3 is 159.7
Fe2O3 + 3H2 ==> 3H2O + 2Fe
mols Fe2O3 = grams/molar mass
Convert mols Fe2O3 to mols Fe using the coefficients.
Convert mols Fe to grams. g = mols x molar mass. This is the theoretical yield.
%yield = (actual yield 2.6g/theore yield)*100 = ?
(4.50g Fe2O3)*(1mol Fe2O3/159.7g Fe2O3)*(111.69g Fe2/1mol Fe2)= 3.15g Fe2
%Yield = (2.60g/3.15g)*100 = 0.825*100 = 82.5%
To find the percent yield, we first need to calculate the theoretical yield and then divide it by the actual yield and multiply by 100.
Step 1: Find the number of moles of Fe2O3.
We can use the molar mass of Fe2O3 to convert grams to moles:
Molar mass of Fe2O3 = 159.7 g/mol
Number of moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3
Number of moles of Fe2O3 = 4.50 g / 159.7 g/mol
Number of moles of Fe2O3 ≈ 0.0282 mol
Step 2: Determine the stoichiometry of the reaction.
From the balanced equation, we can see that 1 mole of Fe2O3 produces 2 moles of Fe:
Fe2O3 + 3H2 → 2Fe + 3H2O
Step 3: Calculate the theoretical yield of Fe.
Theoretical yield of Fe = Number of moles of Fe2O3 × (2 moles of Fe / 1 mole of Fe2O3)
Theoretical yield of Fe = 0.0282 mol × (2 mol / 1 mol)
Theoretical yield of Fe = 0.0564 mol
Step 4: Calculate the percent yield.
Percent yield = (Actual yield / Theoretical yield) × 100
Percent yield = (2.6 g / 0.0564 mol) × 100
Percent yield ≈ 46.1%
Therefore, the percent yield is approximately 46.1%.
To calculate the percent yield, you need to compare the actual yield (in this case, 2.6 g of iron metal) with the theoretical yield (the maximum amount of iron that could be produced based on the reaction stoichiometry).
First, you need to determine the balanced chemical equation for the reaction.
Fe2O3 + 3H2 -> 2Fe + 3H2O
According to the equation, each 1 mole of Fe2O3 reacts with 3 moles of H2 to produce 2 moles of Fe.
Next, convert the mass of Fe2O3 (4.50 g) to moles using its molar mass (159.7 g/mol):
4.50 g Fe2O3 * (1 mol Fe2O3 / 159.7 g Fe2O3) = 0.0282 mol Fe2O3
From the balanced equation, we know that the molar ratio between Fe2O3 and Fe is 1:2. Therefore, the theoretical yield of Fe can be calculated:
0.0282 mol Fe2O3 * (2 mol Fe / 1 mol Fe2O3) = 0.0564 mol Fe
Convert the theoretical yield from moles to grams using the molar mass of Fe (55.8 g/mol):
0.0564 mol Fe * (55.8 g Fe / 1 mol Fe) = 3.14 g Fe
Now that we have the actual yield (2.6 g Fe) and the theoretical yield (3.14 g Fe), we can calculate the percent yield:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Percent Yield = (2.6 g Fe / 3.14 g Fe) * 100 = 82.8%
Therefore, the percent yield of the reaction is approximately 82.8%.