Heating an ore of antimony (Sb2S3) in the presence of iron gives the element antimony and iron(II) sulfide.
Sb2S3(s) + 3 Fe(s) 2 Sb(s) + 3 FeS(s)
When 10.9 g Sb2S3 reacts with an excess of Fe, 9.84 g Sb is produced. What is the percent yield of this reaction?
could you please give me the answer to this problem, im struggling with it. THANKS!!
This same question was posted by Clara. Here is the response I gave.
http://www.jiskha.com/display.cgi?id=1270072207
To calculate the percent yield of a reaction, you need to compare the actual yield (the amount obtained in the experiment) to the theoretical yield (the amount that should have been obtained according to stoichiometry).
1. Calculate the theoretical yield of Sb using stoichiometry:
- From the balanced equation, we can see that the molar ratio between Sb2S3 and Sb is 1:2 (1 mol of Sb2S3 produces 2 mol of Sb).
- Convert the mass of Sb2S3 to moles using its molar mass (Sb2S3: 338.4 g/mol).
- Use the molar ratio to determine the moles of Sb produced.
- Convert moles of Sb to grams using its molar mass (Sb = 121.8 g/mol).
2. Calculate the percent yield:
- Divide the actual yield (9.84 g) by the theoretical yield.
- Multiply the result by 100 to get the percent yield.
Let's calculate it step by step:
Step 1:
Molar mass of Sb2S3 = (2 mol Sb x 121.8 g/mol) + (3 mol S x 32.1 g/mol) = 338.4 g/mol
Moles of Sb2S3 = mass / molar mass = 10.9 g / 338.4 g/mol = 0.032 mol
Moles of Sb = 2 x moles of Sb2S3 = 2 x 0.032 mol = 0.064 mol
Mass of Sb = moles x molar mass = 0.064 mol x 121.8 g/mol = 7.79 g
Step 2:
Percent yield = (actual yield / theoretical yield) x 100
Percent yield = (9.84 g / 7.79 g) x 100 = 126.5%
Therefore, the percent yield of this reaction is 126.5%.