Question

10 g of s reacts with excess of o2 to form 15g of so2 the % yield of this reaction is

Answer 1

S + O2 = SO2

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 %

Answer 2

Answer : The % yield of this reaction is 75 %.

Solution : Given,

Mass of sulfur = 10 g

Mass of $SO_2$ = 15 g

Molar mass of S = 32 g/mole

Molar mass of $SO_2$ = 64 g/mole

The balanced reaction is,

$S+O_2\rightarrow SO_2$

From the balanced reaction, we conclude that

As 1 mole of S produces 1 mole of $SO_2$

32 g of S produces 64 g of $SO_2$

10 g of S produces $\frac{64g}{32g}\times 10g=20g$ of $SO_2$

The actual yield of $SO_2$ = 15 g

The theoretical yield of $SO_2$ = 20 g

Now we have to calculate the % yield of the reaction.

% yield of the reaction = $\frac{\text{ Actual yield}}{\text{ Theoretical yield}}\times 100=\frac{15g}{20}\times 100=75%$

Therefore, the % yield of this reaction is 75 %.