Question #74d98

1 Answer
P dilip_k
Oct 7, 2016

The HCl solution taken 20mL 0.1M
The no. Of moles of HCl in this solution
="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"xx "molarity"
=7.5xx10^(-3)xx0.2=1.5*10^-3" moles"

The equation of the acid base reaction is

HCl+NaOH->NaCl+H_2O

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

So (2*10^-3-1.5*10^-3)=0.5*10^-3moles HCl reacts with Mg

Now equation of the reaction of Mg with HCl is

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*10^-3" mol " HCl will be neutralised by 1/2*0.5*10^-3" mol "Mg

Hence the mass of Mg added to HCl solution was

=1/2*0.5*10^-3" mol"xx24 g/"mol"

=6*10^-3g " "Mg

where atomic mass of Mg =24g/"mol"

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