Answer:
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%