Suppose 87.9 grams of FeCl3 reacts with excess aluminum metal to produce 22.3 grams of Fe. What would be the % yield?
FeCl3 + Al --> AlCl3 + Fe
Why must the water levels be equal in a tube before measuring the volume of gas?
FeCl3 + Al --> AlCl3 + Fe
mols FeCl3 = grams/molar mass.
mols Fe = mols FeCl3 (look at the coefficients in the equation).
Convert mols FeCl3 to g. g = mols x molar mass. This is the theoretical yield.
Then %yield = (actual yield/theoret yield)*100 = ?
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You want the water levels inside the tube and outside the tube to be the same because you want the pressure inside the tube to be the same as that outside the tube. You can't measure the pressure inside the tube (but that's what you want) but you CAN measure the pressure outside the tube (that's atmospheric pressure) and that's easy to measure.
To calculate the percent yield, we need to compare the actual yield with the theoretical yield. The theoretical yield is the amount of product that would be obtained if the reaction proceeded perfectly, assuming all reactants were completely converted into product.
First, let's find the theoretical yield of Fe using stoichiometry. We need to determine the moles of Fe produced from the given mass of FeCl3.
1. Find the molar mass of FeCl3:
Fe: 55.85 g/mol
Cl: 35.45 g/mol (x 3 because there are 3 Cl atoms)
Total molar mass of FeCl3 = 55.85 g/mol + 35.45 g/mol x 3 = 162.2 g/mol
2. Calculate the moles of FeCl3:
moles = mass / molar mass = 87.9 g / 162.2 g/mol ≈ 0.541 mol FeCl3
3. Use the mole ratio between FeCl3 and Fe to determine the moles of Fe produced:
From the balanced equation, we can see that the stoichiometric ratio between FeCl3 and Fe is 1:1. Therefore, moles of Fe produced = moles of FeCl3 = 0.541 mol.
4. Calculate the mass of Fe using the moles and molar mass of Fe:
mass = moles x molar mass = 0.541 mol x 55.85 g/mol ≈ 30.2 g
Now we have the theoretical yield, which is approximately 30.2 grams of Fe.
To find the percent yield:
Percent yield = (actual yield / theoretical yield) x 100
In this case, the actual yield is given as 22.3 grams of Fe.
Percent yield = (22.3 g / 30.2 g) x 100 ≈ 73.8%
Therefore, the percent yield of Fe is about 73.8%.