Heating an ore of antimony (Sb2S3) in the presence of iron yields antimony and iron (II) sulfide.
Sb2S3 + 3 Fe = 2Sb + 3 FeS
When 15g of Sb2S3 reacts with 10g of Fe in the lab 9.84g of Sb is produced.
What is the limiting reactant in this reaction?
What is the theorectical yield of Sb in this reaction?
What is the percent yield?
To determine the limiting reactant, we need to calculate the amount of Sb that would be produced if all of the Fe reacted.
First, we calculate the molar mass of Sb2S3:
Sb = 121.75 g/mol
S = 32.06 g/mol
Molar mass of Sb2S3 = (2 x 121.75 g/mol) + (3 x 32.06 g/mol) = 339.64 g/mol
Next, we calculate the number of moles of Sb2S3 and Fe:
Number of moles of Sb2S3 = mass / molar mass = 15 g / 339.64 g/mol = 0.0442 mol
Number of moles of Fe = mass / molar mass = 10 g / 55.85 g/mol = 0.179 mol
The stoichiometric ratio between Sb2S3 and Fe is 1:3, meaning that for every 1 mole of Sb2S3, 3 moles of Fe are required. Therefore, the number of moles of Sb that would be produced if all of the Fe reacted is 3 times the number of moles of Sb2S3.
Number of moles of Sb = 3 x 0.0442 mol = 0.1326 mol
Now, we can calculate the theoretical yield of Sb:
Theoretical yield of Sb = number of moles of Sb x molar mass of Sb = 0.1326 mol x 121.75 g/mol = 16.152 g
The theoretical yield of Sb is 16.152 g.
Finally, we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) x 100
Given that the actual yield of Sb is 9.84 g, we can substitute these values into the equation:
Percent yield = (9.84 g / 16.152 g) x 100 = 60.85%
Therefore, the limiting reactant in this reaction is Fe. The theoretical yield of Sb is 16.152 g. The percent yield is 60.85%.
To determine the limiting reactant, theoretical yield, and percent yield of the reaction, we need to follow a few steps.
1. Calculate the moles of each reactant:
The molar mass of Sb2S3 is 339.68 g/mol.
The molar mass of Fe is 55.85 g/mol.
The number of moles of Sb2S3:
moles of Sb2S3 = mass of Sb2S3 / molar mass of Sb2S3
moles of Sb2S3 = 15 g / 339.68 g/mol
moles of Sb2S3 ≈ 0.0442 mol
The number of moles of Fe:
moles of Fe = mass of Fe / molar mass of Fe
moles of Fe = 10 g / 55.85 g/mol
moles of Fe ≈ 0.179 mol
2. Determine the stoichiometric ratio:
From the balanced equation, we can see that the ratio of moles of Sb2S3 to Fe is 1:3. This means that for every 1 mole of Sb2S3, we need 3 moles of Fe.
3. Identify the limiting reactant:
To determine the limiting reactant, we compare the mole ratio of the reactants in the balanced equation to the actual mole ratio of the reactants in the experiment.
Mole ratio of Sb2S3 to Fe: 1:3
Mole ratio of Sb2S3 to Fe in the experiment: 0.0442 mol : 0.179 mol
Since the mole ratio is less than the stoichiometric ratio, Fe is the limiting reactant.
4. Calculate the theoretical yield:
The balanced equation tells us that 1 mole of Sb2S3 produces 2 moles of Sb. Therefore, the stoichiometric ratio of Sb2S3 to Sb is 1:2.
moles of Sb = moles of Sb2S3 × (2 moles of Sb / 1 mole of Sb2S3)
moles of Sb = 0.0442 mol × (2 mol / 1 mol)
moles of Sb = 0.0884 mol
mass of Sb = moles of Sb × molar mass of Sb
mass of Sb = 0.0884 mol × 121.8 g/mol (molar mass of Sb)
mass of Sb ≈ 10.77 g
Therefore, the theoretical yield of Sb is 10.77 g.
5. Calculate the percent yield:
Percent yield = (actual yield / theoretical yield) × 100
actual yield of Sb = 9.84 g
Percent yield = (9.84 g / 10.77 g) × 100
Percent yield ≈ 91.27%
So, the limiting reactant in this reaction is Fe.
The theoretical yield of Sb is 10.77 g.
The percent yield is approximately 91.27%.
mols Sb2S3 = grams/molar mass
mols Fe = grams/atomic mass
Convert mols Sb2S3 to mols Sb using the coefficients in the balanced equation.
Convert mols Fe to mols Sb.
Most likely mols Fe will not agree which means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent providing that number is the limiting reagent.
Using the smaller value, convert to g. g = mols Sb x atomic mass Sb. This is the theoretical yield.
%yield = (actual yield/theor yield)*100 = ?