Question

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of 5.0 grams of O2 with 6.0 grams of S. What is the % yield of SO3 in this experiment? S (s) + O2 (g) → SO3 (g) (not balanced)

Answer 1

Answer:

94.7 %

Explanation:

The balanced chemical reaction is as follows -

2 S + 3 O₂ -----> 2 SO₃

Hence ,

The number of moles according to the chemical reaction is -

Moles of S = 2

Moles of O₂ = 3

Since ,

Moles is denoted by given mass divided by the molecular mass ,  

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

From the question ,

For Sulfur ,

w = 6.0 g

as we know the molecular mass of S = 32 ,

Hence ,

moles are calculated by using the above formula , and putting the respective values ,

n = w / m  = 6 / 32 = 0.1871 mol

similarly for Oxygen ,

w = 5.0 g

as we know the molecular mass of O = 32 g/mol ,

Hence ,

moles are calculated by using the above formula , and putting the respective values ,

n = w / m  = 5 / 32 = 0.15625 mol

Comparing the moles with the values from the the equation ,

Since , S is in excess ,

Therefore , O₂ is the limiting reagent ,

Hence O₂ will determine the moles of the product ,

Therefore ,

Using unitary method ,

From the chemical reaction ,

3 mol of O₂ will give 2 mol of SO₃

and ,

1 mol of O₂ will give 2 / 3 mol of SO₃

Therefore ,

0.15625 mol O₂ will give 2 / 3 * 0.15625 mol of SO₃

Hence ,

Moles of  SO₃ produced = 0.1042 mol

since ,

n = w / m

w = n * m

molecular mass of SO₃ = 80 g/mol

w = Mass of SO₃ =  0.1042 mol * 80 g/mol = 8.340 g

The percentage yield is calculated as the actual yield divided by the theoretical yield multiplied by 100 .

Hence ,

Percentage yield = actual yield / theoretical yield * 100

From the question ,

actual yield = 7.9 g of  SO₃

As calculated above ,

theoretical yield = 8.340 g

Percentage yield =  7.9 g / 8.340 g * 100 = 94.7 %