Determine the percent yield for the reaction between 276.94 g of Sb2S3 and excess oxygen if 59.19 g of Sb4O6 is recovered along with an unknown amount of sulfur dioxide gas. (include 2 decimal places in your answer)
I worked this problem, but with different numbers, several years ago. Just follow the steps but substitute the number in your problem. Post your work if you have any problems.
https://www.jiskha.com/questions/81919/determine-the-percent-yield-for-the-reaction-between-98-7g-of-sb2s3-and-excess-oxygen-gas
To determine the percent yield for a reaction, you need to compare the actual yield (the amount of product obtained) to the theoretical yield (the amount of product that should have been obtained according to stoichiometry). The percent yield can be calculated using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) × 100
In this case, the reaction is between Sb2S3 and oxygen, and the product obtained is Sb4O6.
First, we need to calculate the theoretical yield of Sb4O6. To do this, we need to determine the stoichiometry of the reaction from the balanced chemical equation. Let's write the balanced equation for the reaction:
4 Sb2S3 + 13 O2 -> 8 Sb4O6 + 6 SO2
From the balanced equation, we can see that for every 4 moles of Sb2S3, 8 moles of Sb4O6 should be produced.
Now, we need to convert the given mass of Sb2S3 (276.94 g) to moles. The molar mass of Sb2S3 is:
2 * (121.76 g/mol for antimony) + 3 * (32.06 g/mol for sulfur) = 339.64 g/mol
So, the moles of Sb2S3 can be calculated as:
(mass of Sb2S3 / molar mass of Sb2S3) = (276.94 g / 339.64 g/mol) = 0.815 mol
According to stoichiometry, the theoretical yield of Sb4O6 can be calculated as:
(theoretical yield of Sb4O6) = (moles of Sb2S3) * (moles of Sb4O6 / moles of Sb2S3) = 0.815 mol * (8 mol / 4 mol) = 1.63 mol
Next, we need to find the actual yield of Sb4O6, which is given as 59.19 g.
Now, we can calculate the percent yield using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) × 100
Percent Yield = (59.19 g / (1.63 mol * (334.72 g/mol for Sb4O6))) × 100
Percent Yield = (59.19 g / 539.78 g) × 100
Percent Yield ≈ 10.94%
Therefore, the percent yield for the reaction is approximately 10.94%.