Question #74d98

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Question #74d98

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Answer 1 · 2016-10-07T08:46:36

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

The NaOH solution required for neutralisation of excess acid is 7.5mL 0.2M
The no. Of moles of NaOH in this solution
$= volume in L \times molarity$ $="volume in L"xx "molarity"$
$= 7.5 \times 10^{- 3} \times 0.2 = 1.5 \cdot 10^{- 3} moles$ $=7.5xx10^(-3)xx0.2=1.5*10^-3" moles"$

The equation of the acid base reaction is

$H C l + N a O H \to N a C l + H_{2} O$ $HCl+NaOH->NaCl+H_2O$

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

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

Now equation of the reaction of Mg with HCl is

$2 H C l + M g \to M g C l_{2} + H_{2}$ $2HCl+Mg->MgCl_2+H_2$

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

So

$0.5 \cdot 10^{- 3} mol H C l$ $0.5*10^-3" mol " HCl$ will be neutralised by $\frac{1}{2} \cdot 0.5 \cdot 10^{- 3} mol M g$ $1/2*0.5*10^-3" mol "Mg$

Hence the mass of Mg added to HCl solution was

$= \frac{1}{2} \cdot 0.5 \cdot 10^{- 3} mol \times 24 \frac{g}{mol}$ $=1/2*0.5*10^-3" mol"xx24 g/"mol"$

$= 6 \cdot 10^{- 3} g M g$ $=6*10^-3g " "Mg$

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

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Question #74d98

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