Question

For the balanced equation shown below, if the reaction of 24.3 grams of SO2 react with 42.3 g of H2S producing 4.58 grams of sulfur, what is the percent yield?
   2 H2S + SO2 → 3 S + 2 H2O

Answer 1

Answer:

2 H₂S + SO₂ → 3 S + 2 H₂O

SO₂ wieght = 24.3 g       SO₂ molar mass = 64

SO₂ moles = weight / molar mass = 24.3 / 64

                                                        = 0.379

H2S moles :

H₂S weight = 42.3 g

H₂S molar mass = 34

moles = 42.3 / 34

= 1.244

S moles:

S weight = 4.58 g (produced in the reaction)

S molar mass = 32

moles = 4.58 / 32

= 0.143

2 H₂S + SO₂  → 3 S + 2 H₂O

2 1 3 moles from the equation

1.243 0.379 0.143

In the reaction SO₂ moles is less. Hence it is the limiting factor of the reaction (Yield caluculated based on this)

2 H₂S + SO₂  → 3 S + 2 H₂O

64 3 x 32 From molar mass

24.3....... ? weight from question

= (24.3 x 3 x 32) / 64

= 2332.8 / 64

= 36.45

Hence the 100% yield = 36.45

Produced yield in the reaction =4.58 g

Yield caluculation:

36.45 ........... 100%

4.58 .................... ?

= (4.58 x 100) / 36.45

= 12.565

Hence the percentage of the yield = 12.565