Question

4gm mixture of Na2Co3 and NaHCo3 on heating liberates 448ml of Co2 at STP. the percentage of Na2Co3 in mixture is?​

Answer 1

Answer : The percentage of $Na_2CO_3$ in mixture is, 16%

Explanation : Given,

Mass of mixture = 4 g

Volume of $CO_2$ = 448 ml = 0.448 L

Molar mass of $NaHCO_3$ = 84 g/mole

First we have to calculate the moles of $CO_2$.

At STP,

As, 22.4 L volume of $CO_2$ present in 1 mole of $CO_2$

As, 0.448 L volume of $CO_2$ present in $\frac{0.448}{22.4}=0.02mole$ of $CO_2$

Now we have to calculate the moles of $Na_2CO_3$.

As we know that $Na_2CO_3$ is a strong base and it will not decompose on normal heating.

The decomposition of $NaHCO_3$ will be :

$2NaHCO_3\rightarrow Na_2CO_3+CO_2+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CO_2$ obtained from 2 moles of $NaHCO_3$

So, 0.02 mole of $CO_2$ obtained from 0.04 moles of $NaHCO_3$

Now we have to calculate the mass of $NaHCO_3$.

$\text{Mass of }NaHCO_3=\text{Moles of }NaHCO_3\times \text{Molar mass of }NaHCO_3=0.04mole\times 84g/mole=3.36g$

Now we have to calculate the mass of $Na_2CO_3$.

The mass of $Na_2CO_3$ = 4 - 3.36 = 0.64 g

Now we have to calculate the percentage of $Na_2CO_3$ in the mixture.

$\%\text{ of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Total mass}}\times 100=\frac{0.64g}{4g}\times 100=16\%$

Therefore, the percentage of $Na_2CO_3$ in mixture is, 16%

Answer 2

Answer:

Answer : The percentage of Na_2CO_3Na

2

CO

3

in mixture is, 16%

Explanation : Given,

Mass of mixture = 4 g

Volume of CO_2CO

2

= 448 ml = 0.448 L

Molar mass of NaHCO_3NaHCO

3

= 84 g/mole

First we have to calculate the moles of CO_2CO

2

.

At STP,

As, 22.4 L volume of CO_2CO

2

present in 1 mole of CO_2CO

2

As, 0.448 L volume of CO_2CO

2

present in \frac{0.448}{22.4}=0.02mole

22.4

0.448

=0.02mole of CO_2CO

2

Now we have to calculate the moles of Na_2CO_3Na

2

CO

3

.

As we know that Na_2CO_3Na

2

CO

3

is a strong base and it will not decompose on normal heating.

The decomposition of NaHCO_3NaHCO

3

will be :

2NaHCO_3\rightarrow Na_2CO_3+CO_2+H_2O2NaHCO

3

→Na

2

CO

3

+CO

2

+H

2

O

From the balanced reaction we conclude that,

As, 1 mole of CO_2CO

2

obtained from 2 moles of NaHCO_3NaHCO

3

So, 0.02 mole of CO_2CO

2

obtained from 0.04 moles of NaHCO_3NaHCO

3

Now we have to calculate the mass of NaHCO_3NaHCO

3

.

\text{Mass of }NaHCO_3=\text{Moles of }NaHCO_3\times \text{Molar mass of }NaHCO_3=0.04mole\times 84g/mole=3.36gMass of NaHCO

3

=Moles of NaHCO

3

×Molar mass of NaHCO

3

=0.04mole×84g/mole=3.36g

Now we have to calculate the mass of Na_2CO_3Na

2

CO

3

.

The mass of Na_2CO_3Na

2

CO

3

= 4 - 3.36 = 0.64 g

Now we have to calculate the percentage of Na_2CO_3Na

2

CO

3

in the mixture.

\%\text{ of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Total mass}}\times 100=\frac{0.64g}{4g}\times 100=16\%% of Na

2

CO

3

=

Total mass

Mass of Na

2

CO

3

×100=

4g

0.64g

×100=16%

Therefore, the percentage of Na_2CO_3Na

2

CO

3

in mixture is, 16%