Question

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

Answer 1

2Sb2S3 + 9O2 ==> Sb4O6+ 6SO2

Convert 98.7 g Sb2S3 to mols.
Convert mols Sb2S3 to mols Sb4O6 using the coefficients in the balanced equation.
Convert mols Sb4O6 to grams Sb4O6. (This is the theoretical yield.)
Convert grams Sb4O6 to percent Sb4O6.
%Sb4O6 = [72.4/theoretical yield]*100
Post your work if you get stuck.

Answer 2

72.4/97.82*100?

Answer 3

I don't get 97.82 but more like 62 something.

Answer 4

85.5%

Answer 5

98.7g Sb2S3 | 1mol Sb2S3 | 1mol Sb4O6 |
-----------------|---------------------|------------------|-->
| 333.9g Sb2S3 | 2mol Sb2S3 |
| 583.3g Sb4O6 | 57571.71 |
>|----------------------|=----------------|= 84.7g
| 1mol Sb4O6 | 679.8 | Sb4O6

72.4g Sb4O6
------------------x100=85.5%
84.7g Sb4O6

Answer 6

Determine the percent yield for the reaction between 162 g of Sb2S3 and excess oxygen if 55 g of Sb4O6 is recovered along with an unknown amount of sulfur dioxide gas?

Answer 7

85.5%