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 %
