Heating an ore of antimony (Sb2S3) in the presence of iron gives the element antimony and iron(II) sulfide.
Sb2S3(s) + 3Fe(s) ==> 2Sb(s) + 3FeS(s)
When 15.0g Sb2S3 reacts with an excess of Fe, 9.84g Sb is produced. What is the percent yield of this reaction?
1. You have the equation.
2. Convert 15.0 g Sb2S3 to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles Sb2S3 to moles Sb.
4. Now convert moles Sb to grams. g = moles x molar mass. This is the theoretical yield.
5. %yield = (actual yield/theoretical yield)*100 = ??
The problem states that 9.84 g is the actual yield.
55.9
To calculate the percent yield of a reaction, you need to compare the actual yield with the theoretical yield.
1. Calculate the theoretical yield:
- Convert the mass of Sb2S3 to moles using its molar mass.
Molar mass of Sb2S3 = 2(121.8 g/mol) + 3(32.1 g/mol) = 339.4 g/mol
Moles of Sb2S3 = 15.0 g / 339.4 g/mol = 0.0442 mol
- Use the balanced equation to determine the molar ratio between Sb2S3 and Sb.
According to the equation: 1 mol of Sb2S3 produces 2 mol of Sb.
Moles of Sb produced = 2 * 0.0442 mol = 0.0884 mol
- Convert the moles of Sb to grams using its molar mass.
Molar mass of Sb = 121.8 g/mol
Theoretical mass of Sb = 0.0884 mol * 121.8 g/mol = 10.77 g
2. Calculate the percent yield:
- Percent yield = (Actual yield / Theoretical yield) * 100
Actual yield = 9.84 g
Percent yield = (9.84 g / 10.77 g) * 100 = 91.3%
Therefore, the percent yield of this reaction is approximately 91.3%.
To find the percent yield of the reaction, you need to compare the actual yield (the amount of Sb obtained in the reaction) to the theoretical yield (the maximum amount of Sb that can be obtained based on the balanced equation).
First, let's calculate the theoretical yield of Sb using the given information:
1) Calculate the molar mass of Sb2S3:
Sb2S3: (2 x molar mass of Sb) + (3 x molar mass of S)
Sb2S3: (2 x 121.75 g/mol) + (3 x 32.07 g/mol)
Sb2S3: 243.50 g/mol + 96.21 g/mol
Sb2S3: 339.71 g/mol
2) Use the molar ratio from the balanced equation to convert the mass of Sb2S3 to moles of Sb:
(2 moles of Sb / 1 mole of Sb2S3) x (15.0 g Sb2S3 / 339.71 g/mol Sb2S3)
moles of Sb = (2/1) x (15.0/339.71) ≈ 0.088 mol Sb
3) Calculate the molar mass of Sb:
Molar mass of Sb: 121.75 g/mol
4) Calculate the theoretical yield of Sb:
theoretical yield = moles of Sb x molar mass of Sb
theoretical yield = 0.088 mol Sb x 121.75 g/mol
theoretical yield ≈ 10.71 g Sb
Now, let's calculate the percent yield:
5) percent yield = (actual yield / theoretical yield) x 100%
percent yield = (9.84 g / 10.71 g) x 100%
percent yield ≈ 91.8%
Therefore, the percent yield of this reaction is approximately 91.8%.