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 15.0 g Sb2Sb3 reacts with an excess of Fe, 9.84 g Sb is produced. What is the percent yield of this reaction?
Here is a worked example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the percent yield of a reaction, you need to compare the actual yield (the amount of product obtained in the experiment) with the theoretical yield (the amount of product that should have been obtained according to stoichiometry).
First, let's calculate the molar mass of Sb2S3 and Sb:
- Molar mass of Sb2S3 = (2 * atomic mass of Sb) + (3 * atomic mass of S) = (2 * 121.75 g/mol) + (3 * 32.06 g/mol) = 339.64 g/mol
- Molar mass of Sb = 121.75 g/mol
Next, we'll use the molar mass to convert the given masses of Sb2S3 and Sb into moles:
- Moles of Sb2S3 = 15.0 g / 339.64 g/mol = 0.0441 mol
- Moles of Sb = 9.84 g / 121.75 g/mol = 0.0807 mol
According to the balanced chemical equation for the reaction, the molar ratio between Sb2S3 and Sb is 1:2. This means that for every 1 mole of Sb2S3, 2 moles of Sb should be produced.
Now, let's calculate the theoretical yield of Sb:
- Theoretical yield of Sb = (0.0441 mol Sb2S3) x (2 mol Sb / 1 mol Sb2S3) x (121.75 g/mol Sb) = 10.90 g
Finally, we can calculate the percent yield using the formula:
- Percent yield = (Actual yield / Theoretical yield) x 100
- Percent yield = (9.84 g / 10.90 g) x 100 = 90.37%
Therefore, the percent yield of this reaction is approximately 90.37%.