Question

pie 8: 40 g mixture of Na,co, and Caco, is gently heated. On heating this sample produces 2.24 litre of Co, at STP. Calculate the % weight of Caco, in that sample. We now that Na co does not decompose on heating. So CO, will be produced by only Caco,​

Answer 1

Answer:

The weight % of $CaCO_3$  in the sample that contains $40\ g$ mixture of $Na_2CO_3$ and $CaCO_3$  is $25$ %·

Explanation:

$Na_2CO_3$ does not decompose on heating· Only $CaCO_3$ decompose on heating and produce the carbon dioxide·

The decomposing reaction of  $CaCO_3$ follows,

$CaCO_3\ --->\ CaO\ +\ CO_2$

Given that,

The weight of the mixture  $($$Na_2CO_3$ $+$$CaCO_3$$)\ =$ $40\ g$

The volume of $CO_2$ produced at STP $=$ $2.24\ L$

To calculate the weight  %  of $CaCO_3$  in the sample first, we need to find out the weight of $CaCO_3$ present in the sample·

For that, let us take the weight of  $CaCO_3$  in the mixture as X g·

Then,

From the reaction it is clear that,

One mole of  $CaCO_3$  decomposes to give one mole of CaO and one mole of $CO_2$·

So,

One mole of $CaCO_3$  $=\ \frac{weight\ of\ CaCO_3 }{Molar\ mass\ of\ CaCO_3}$

The weight of $CaCO_3$  in the mixture we took it as X g·

Molar mass of  $CaCO_3$  $=\ 100\ g$

On substituting the values we get,

One mole of   $CaCO_3$  $=\ \ \frac{X\ g}{100\ g}$

It is clear that,

One mole of  $CaCO_3$  produces $2.24\ L$ of  $CO_2$ at STP·

We know that,

One mol of $CO_2$ gas will occupy $22.4\ L$ at STP·

Then,

$2.24\ L$ of  $CO_2$ gas will have $\frac{2.24\ L}{22.4\ L}$ moles of  $CO_2$ ·

That is,

$0.1$ mol of $CO_2$ gas·

So,

$\frac{X}{100}$ mol  of  $CaCO_3$  produces $0.1$ mol of $CO_2$ gas·

Then,

$X\ \ =\ \ 10g$

Hence,

The weight of  $CaCO_3$  in the sample  $=$  $X\ \ =\ \ 10g$

Therefore,

The weight %  $=\ \frac{Weight\ Of\ CaCO_3}{Total\ Weight\ Of\ The\ Sample}$   ×  $100$

Weight of   $CaCO_3$  $=\ 10\ g$

Total weight of the sample $=\ 40\ g$

On substituting the values we get,

weight % of   $CaCO_3$ $=\ \frac{10\ g}{40\ g} \ *\ 100$

weight % of   $CaCO_3$ $=\ 25$ %

Hence,

The weight % of  $CaCO_3$ in the sample that contains $40\ g$ mixture of $Na_2CO_3$ and $CaCO_3$  is $25$ %·