Question

Does anybody know how to do q4. Please show working out thanks.

Answer 1

% purity of limestone = 96.53%

Step-by-step explanation:

Question (4).

Weight of impure CaCO₃ = 25.9 g

Molecular weight of CaCO₃ = 40 + 12 + 3(16)

= 100 g per mole

We know at S.T.P. number of moles of CO₂ = 1 and volume = 22.4 liters

From the given reaction, 1 mole of CaCO₃ reacts with 1 mole or 22.4 liters of

CO₂.

∵ 22.4 liters of CO₂ was produced from CaCO3 = 100 g

∴ 1 liter of CO₂ will be produced by CaCO₃ =
`(100)/(22.4)`

∴ 5.6 liters of CO₂ will be produced by CaCO₃ =
`(100* 5.6)/(22.4)`

= 25 g

Therefore, % purity of CaCO₃ =
`\frac{\text{Weight calculated}}{{\text{Weight given}}}* 100`

=
`(25)/(25.9)* 100`

= 96.53 %