Question

Determine the percent yield for the reaction between 275.37 g of Sb2S3 and excess oxygen if 59.65 g of Sb4O6 is recovered along with an unknown amount of sulfur dioxide gas. (include 2 decimal places in your answer)

Answer 1

Answer:

25.23%

Explanation:

The reaction that takes place is:

• 2Sb₂S₃ + 9O₂ → Sb₄O₆ + 6SO₂

Sb₂S₃ molecular weight = 121.76*2 + 32*3 = 339.52 g/mol

Sb₄O₆ molecular weight = 121.76*4 + 16*6 = 583.04 g/mol

First we calculate the mass of Sb₄O₆ that would be produced if the yield was 100%:

• 275.37 gSb₂S₃ ÷ 339.52 g/mol * $\frac{1molSb_{4}O_{6}}{2molSb_{2}S_{3}}$ * 583.04 g/mol = 236.44 g Sb₄O₆

Now we calculate the yield, using the mass of Sb₄O₆ that was actually produced:

• 59.65 g / 236.44 g * 100% = 25.23%