Question

What volume of Sulphur dioxide (at s.t.p) would be liberated by roasting 30 g of pyrites?

Answer 1

Step 1:Write the chemical reaction

$4{\mathrm{FeS}}_2+11{\mathrm{O}}_2\to 2{\mathrm{Fe}}_2{\mathrm{O}}_3+8{\mathrm{SO}}_2$

Iron disulphide oxygen Iron(III) oxide Sulphur dioxide

Given:

(S= 32, Fe= 56, molar volume of gas is 22.4 litres at s.t.p)

Step 2: Find the molar mass

Molar mass of ${\mathrm{FeS}}_2$$=56+2x32$

$=56+64$

$=120g/mol$

According to the equation, the molecular weight of 4 moles of ${\mathrm{FeS}}_2=4\mathrm{x}120$

$=480g/mol$

The molar volume of gas is 22.4 litres at s.t.p

So, the molar volume of $8{\mathrm{SO}}_2$$=8x22.4litres$

$=179.2litres$

Step 3 Find the volume:

480 g of ${\mathrm{FeS}}_2$ on roasting gives 179.2 litres of ${\mathrm{SO}}_2$ at S.T.P.

Therefore, 30 g of ${\mathrm{FeS}}_2$ on roasting will give $=179.2x\frac{30}{480}=11.2litres$ of ${\mathrm{SO}}_2$ at S.T.P.

Step 4: Result

Hence, 11.2 litres of the volume of Sulphur dioxide (at s.t.p) would be liberated.