Discussion Forum

28 g KOH is required to completely neutralise $CO_{2}$ produced on heating 60 g of imnpure $CaCo_{3}$ .The percentage  purity  of $CaCO_{3}$ is approximately  (molar mass of KOH and $CaCO_{3}$  are 56 and 100 g mol^-1, repectively)

Answer:

Explanation:

The  given relation  $CaCO_{3}$  $\underrightarrow{2KOH}$   $CO_{2}$

                                                          $\triangle$

means, 

2 moles of KOH will neutralise 1 mole of  $CO_{2}$ , given

 (i)  KOH (used)=28g

(ii) Molar mass of KOH (M=56g

 (iii) MOlar mass of $CaCO_{3}$ =100g

(iv)  Impure $CaCO_{3}$=60g

 $\because$  112g(56x2) of KOH wil neutralise=100g of $CaCO_{3}$

$\therefore$  28g KOH will neutralise= $\frac{100 \times 28}{112}$

=25g of $CaCO_{3}$

Also,

$\because$ 60g(impure) of $CaCO_{3}$. has 25g pure $CaCO_{3}$,

$\therefore$ 100g(impure) $CaCO_{3}$ has pure $CaCO_{3}$

  =$\frac{25 \times 100}{60}=41.6g$