Question

Determine the percent yield for the reaction between 98.7g of Sb2S3 and excess oxygen gas if 72.4g of Sb4O6 is recovered along with an unknown amount of sulfur dioxide?

Answer 1

the percent yield of Sb4O6 in this reaction is 14.9%

Step-by-step explanation:

The percent yield of a reaction is the actual yield of a product obtained in a reaction divided by the theoretical yield, multiplied by 100%. The theoretical yield is the amount of product that should be obtained based on stoichiometric calculations assuming complete conversion of the limiting reagent.

In this reaction, the balanced equation is:

2 Sb2S3 + 9 O2 → 2 Sb4O6 + 6 SO2

To determine the theoretical yield of Sb4O6, we need to first determine the limiting reagent. The molar masses of Sb2S3 and O2 are:

Sb2S3: 2 x 121.8 g/mol (2 Sb atoms x 121.8 g/mol + 3 S atoms x 32.1 g/mol) = 243.6 g/mol

O2: 2 x 16.0 g/mol = 32.0 g/mol

Using the given mass of Sb2S3 (98.7 g) and the molar mass of Sb2S3, we can calculate the number of moles of Sb2S3:

98.7 g / 243.6 g/mol = 0.405 mol

From the balanced equation, we see that 2 moles of Sb2S3 produce 2 moles of Sb4O6. So the theoretical yield of Sb4O6 is:

0.405 mol Sb2S3 x (2 mol Sb4O6 / 2 mol Sb2S3) x (597.9 g Sb4O6 / mol) = 486.3 g Sb4O6

Now we need to calculate the actual yield of Sb4O6. The given mass is 72.4 g.

Finally, we can calculate the percent yield:

Percent yield = (actual yield / theoretical yield) x 100%

Percent yield = (72.4 g / 486.3 g) x 100% = 14.9%

Therefore, the percent yield of Sb4O6 in this reaction is 14.9%