# 2.6    How many mL of $\inline 0.1\; M\; HCl$ are required to react completely with $\inline 1 \; g$ mixture of $\inline Na_{2}CO_{3}$  and $\inline NaHCO_{3}$  containing equimolar amounts of both?

Total amount of mixture of Na2CO3 and NaHCO3 = 1 g.

Let the amount of Na2CO3 be x g.

So the amount of NaHCO will be equal to (1 - x) g.

$Molar\ mass\ of\ Na_2CO_3 = 106\ ;\ molar\ mass\ of\ NaHCO_3 = 84$

Now it is given that it is an equimolar mixture.

So,            Moles of Na2CO3 =  Moles of  NaHCO3.

or                          $\frac{x}{106} = \frac{1-x}{84}$

or                            x = 0.558 g

So                       $Moles\ of \ Na_2CO_3 = \frac{0.558}{106} = 0.00526$

and                     $Moles\ of \ NaHCO_3 = \frac{1 - 0.558}{84} = 0.0053$

It is clear that for 1 mol of Na2CO3  2 mol of HCl is required, similarly for 1 mol of NaHCO3 1 mol of HCl is required.

So number of moles required of HCl =  2(0.00526) + 0.0053 = 0.01578 mol

It is given that molarity of HCl is 0.1 which means 0.1 mol of HCl in 1l of solution.

Thus required volume :

$= \frac{0.01578}{0.1} = 0.1578\ l = 157.8\ mL$

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