Question

10g of S react with excess of O2 to form 15 g of SO2 the % yield of reaction is

Answer 1

Theoretical yield

The amount of product that could possibly be produced in a given reaction, calculated according to the starting amount of the limiting reagent.

Actual yield

The amount of product actually obtained in a chemical reaction.

% yield is calculated as follows:

(actual yield / theoretical yield )×100

Calculate the actual theoretical yield of the reaction: 

S₂ + 2 O₂ = 2 SO₂

Moles of S₂moles = mass/molar mass        = 10/64       = 0.15625

Mole ratio between S:SO2  = 1:2

That means the moles of SO₂ = 2 x 0.15625 = 0.3125

Find theoretical mass of the SO₂

Mass = moles × molar mass         
          = 0.3125 × 64          = 20 g
Therefore:

Actual yield = 15g

Theoretical yield = 20g

% yield = 15 g /20 g  × 100%             
             = 0.75 × 100%             
              =   75%
The % yield  = 75%

Answer 2

Answer:

S + 2O2 = 2SO2

32 g sulphur yields 64 g SO2

10 g Sulphur will yield 64 x 10 /32 = 20 g

But the actual yield is 15 g

So the percentage yield = 15 x 100/20 = 75 %