Question

20 gms of sulphur on burning in air produces 11.2 ltrs of SO2 at STP the percentage of unreacted sulphur is (a):80 (b):20 (c):60 (d): 40

Answer 1

I think this will help you

Answer 2

Answer: The correct answer is Option b.

Explanation:

STP conditions:

1 mole of a gas occupies 22.4 L of volume.

We know that:

Molar mass of sulfur = 32 g/mol

The chemical equation for the formation of sulfur dioxide from sulfur follows:

$S+O_2\rightarrow SO_2$

By Stoichiometry of the reaction:

If, 22.4 L of volume of sulfur dioxide is produced when 32 g of sulfur is burnt.

So, 11.2 L of volume of sulfur dioxide will be produced when = $\frac{32}{22.4}\times 11.2=16g$ of sulfur is burnt.

Amount of sulfur unreacted = Total amount - amount reacted

Amount of sulfur unreacted = 20 - 16 = 4 g

To calculate the percentage of sulfur unreacted, we use the equation:

$\%\text{ of sulfur unreacted}=\frac{\text{Amount unreacted}}{\text{Total amount}}\times 100$

Amount unreacted = 4 g

Total amount = 20 g

Putting values in above equation, we get:

$\%\text{ of sulfur unreacted}=\frac{4g}{20g}\times 100=20\%$

Hence, the correct answer is Option b.