Question

Question #74d98

Answer 1

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^-3`moles 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"`