Question

Illustration 1: When a mixture of NaBr and NaCl is repeatedly digested with H,SO, all the halogens
are expelled and Na, So, is formed quantitatively. With a particular mixture, it was found
that the weight of Na2SO4 obtained was precisely the same as the weight of Neer-
mixture taken. Calculate the ratio of the weights of NaCl and NaBr in the mixture​

Answer 1

Answer:

NaBr

H

2

SO

4

Na

2

SO

4

a g

NaCl

H

2

SO

4

Na

2

SO

4

b g

∴ Meq. of NaBr+ Meq. of NaCl = Meq. of Na

2

SO

4

Meq. of Na

2

SO

4

formed =

103

a

×1000+

58.5

b

×1000 =9.71a+17.09b

2

142

w

×1000=9.71a+17.09

w=

2000

142

[9.71a+17.09b]

Also,

mass of NaCl + mass of NaBr = mass of Na

2

SO

4

formed

a+b=

2000

142

[9.71a+17.09b]

b

a

=

621.18

426.78

=0.687

Ratio of

NaBr

NaCl

=

a

b

=

1

1.455

So, the ratio as nearest integer is 1

Answer 2

Given:

Mass of $NaBr$ and ​$NaCl$ is equal to the mass of $Na_{2}SO_{4}$

To Find:

Ratio of weight of ​$NaCl$ & $NaBr$ = ?

Solution:

Reaction

$NaBr$ + $H_{2} SO_{4}$ → $Na_{2}SO_{4}$  

​$NaCl$  + $H_{2} SO_{4}$ → $Na_{2}SO_{4}$

Mass of  $NaBr$  = x

Mass of ​$NaCl$   = y  

Mass of $Na_{2}SO_{4}$ = z

Milli-eq of  $NaBr$ + Milli-eq of  ​$NaCl$ = Milli-eq of  $Na_{2}SO_{4}$

Milli-eq. of  $Na_{2}SO_{4}$ =  $\dfrac{x}{103}1000$ +   $\dfrac{y}{58.5}1000$

$\dfrac{z}{142} 2000$    =9.71x + 17.09y

z =  142/ 2000 [9.71x + 17.09y]

As it is given-

Mass of  ​$NaCl$ + Mass of  $NaBr$ = Mass of  $Na_{2}SO_{4}$

x  +  y  =   142/ 2000 [9.71x + 17.09y]

$\dfrac{x}{y} = \dfrac{426.7}{621.1}$

Ratio of  ​$NaCl$ :  $NaBr$ = 621.1 : 426.7