Sb2S3(s) + 3 Fe(s) 2 Sb(s) + 3 FeS(s)
When 14.7 g Sb2S3 reacts with an excess of Fe, 9.84 g Sb is produced. What is the percent yield of this reaction?
%
Follow the steps in this example stoichiometry problem. This will give you the theoretical yield. Then percent yield = (actual yield/theoretical yield)*100 = ?
To calculate the percent yield of a reaction, we need to compare the actual yield (the amount of the desired product obtained experimentally) to the theoretical yield (the maximum amount of product that could have been obtained according to stoichiometry).
In this case, we are given that 14.7 g of Sb2S3 reacts to produce 9.84 g of Sb.
To find the theoretical yield of Sb, we need to use stoichiometry to calculate the amount of Sb that should be produced when 14.7 g of Sb2S3 reacts.
1. Start by determining the molar mass of Sb2S3 and Sb:
- Molar mass of Sb2S3 = (121.76 g/mol)*(2) + (32.07 g/mol)*(3) = 339.63 g/mol
- Molar mass of Sb = 121.76 g/mol
2. Convert the given mass of Sb2S3 to moles:
- Moles of Sb2S3 = 14.7 g / 339.63 g/mol = 0.0433 mol
3. Use stoichiometry to determine the moles of Sb produced:
- From the balanced equation, we see that the molar ratio between Sb2S3 and Sb is 2:2, meaning that for every 2 moles of Sb2S3, 2 moles of Sb are produced.
- Moles of Sb = 0.0433 mol * (2 mol Sb / 2 mol Sb2S3) = 0.0433 mol
4. Convert the moles of Sb to grams:
- Grams of Sb = 0.0433 mol * 121.76 g/mol = 5.27 g
Now we have the theoretical yield of Sb, which is 5.27 g.
To calculate the percent yield, we use the formula:
Percent yield = (actual yield / theoretical yield) x 100
In this case, the actual yield is given as 9.84 g, and the theoretical yield is calculated as 5.27 g:
Percent yield = (9.84 g / 5.27 g) x 100 = 186.6%
Therefore, the percent yield of this reaction is 186.6%.