# Question #74d98

Oct 7, 2016

The HCl solution taken 20mL 0.1M
The no. Of moles of HCl in this solution
$= \text{volume in L"xx "molarity"=20xx10^(-3)xx0.1=2*10^-3" moles}$

The NaOH solution required for neutralisation of excess acid is 7.5mL 0.2M
The no. Of moles of NaOH in this solution
$= \text{volume in L"xx "molarity}$
$= 7.5 \times {10}^{- 3} \times 0.2 = 1.5 \cdot {10}^{-} 3 \text{ moles}$

The equation of the acid base reaction is

$H C l + N a O H \to N a C l + {H}_{2} O$

This equation reveals that acid and base neutralises in 1:1 mole ratio.

So $\left(2 \cdot {10}^{-} 3 - 1.5 \cdot {10}^{-} 3\right) = 0.5 \cdot {10}^{-} 3$moles HCl reacts with Mg

Now equation of the reaction of Mg with HCl is

$2 H C l + M g \to M g C {l}_{2} + {H}_{2}$

This equation reveals that HCl and Mg reacts completely with each other in mole ratio 2:1.

So

$0.5 \cdot {10}^{-} 3 \text{ mol } H C l$ will be neutralised by $\frac{1}{2} \cdot 0.5 \cdot {10}^{-} 3 \text{ mol } M g$

Hence the mass of Mg added to HCl solution was

$= \frac{1}{2} \cdot 0.5 \cdot {10}^{-} 3 \text{ mol"xx24 g/"mol}$

$= 6 \cdot {10}^{-} 3 g \text{ } M g$

where atomic mass of Mg $= 24 \frac{g}{\text{mol}}$