When 20.1 g Sb2S3 reacts with an excess of Fe, 9.84 g Sb is produced. What is the percent yield of this reaction?
My work so far...
converted 20.1g of Sb2S3 to moles
20.1/149.8=.134
.134Sb2S3*2moles Sb/1 mole Sb2S3=.268*121.7=32.6 of Sb <---Thoretical yeild.
9.84/38.6= .3018*100= 30.18%
I have a webassign which tells you if the answer is correct or not and apparently it said mine wasn't
You didn't write the equation but it appears you used Fe2S3 as the product. I think perhaps it is FeS that forms
Sb2S3 + 3Fe ==> 2Sb + 3FeS
Try that and see if it is any better. Your process looks ok if you started with the right equation.
Your calculations seem to be correct. However, there might be a small error in your final calculation of the percent yield.
Let's go through the calculations again to confirm:
1. Convert 20.1 g of Sb2S3 to moles:
20.1 g / 149.8 g/mol = 0.134 mol Sb2S3
2. Use the stoichiometry of the balanced equation to determine moles of Sb produced:
0.134 mol Sb2S3 × (2 mol Sb / 1 mol Sb2S3) = 0.268 mol Sb
3. Calculate the theoretical yield of Sb in grams:
0.268 mol Sb × 121.7 g/mol = 32.6 g Sb (as you correctly found)
4. Calculate the percent yield:
(Actual Yield / Theoretical Yield) × 100
(9.84 g / 32.6 g) × 100 = 30.18%
According to your calculation, the percent yield should be 30.18%. It appears to be correct. However, if your answer was marked incorrect, there might be a mistake in the submission or grading process. Double-check your answer submission and the correct format required by the assignment.
To find the percent yield of a reaction, you need to compare the actual yield (9.84 g of Sb) with the theoretical yield (32.6 g of Sb) and express it as a percentage.
From your calculations, it seems like you made a mistake in determining the theoretical yield. You correctly converted the mass of Sb2S3 to moles, but you need to consider the stoichiometry of the balanced equation to calculate the moles of Sb produced.
The balanced equation for the reaction is:
3Fe + Sb2S3 -> 2Sb + 3FeS
According to the balanced equation, for every 2 moles of Sb2S3 reacted, 2 moles of Sb are produced. Therefore, the moles of Sb produced can be calculated as:
0.134 moles of Sb2S3 * (2 moles of Sb / 2 moles of Sb2S3) = 0.134 moles of Sb
Now, to determine the theoretical yield of Sb, we can multiply the moles of Sb by its molar mass:
0.134 moles of Sb * 121.7 g/mol = 16.34 g of Sb (theoretical yield)
Now, to calculate the percent yield, we divide the actual yield by the theoretical yield and multiply by 100:
Percent yield = (9.84 g / 16.34 g) * 100 = 60.18%
Therefore, the correct percent yield of this reaction is 60.18%. It seems that you made an error in calculating the theoretical yield, resulting in an incorrect answer.