Calculate the mas of Magnesium required to completely react with `250cm^(3)` of 0.1 MHCl. Atomic mass of `Mg=24`
Or (Internal choice)
`100cm^(3)` of a solution of HCl completely neutralizes `25cm^(3)` of 0.1 M NaOH.
Calculate the mass of HCl present in `100cm^(3)`.

Explanation

`250xx0.1M=25` millimoles of HCl
From the equation it is clear that 2HCl is reacting with one Mg.
`:.` 25 millimoles of HCl react with 12.5 millimoles of Mg
`:.` Mass of magnesium required `=` volume of Mg `xx` atomic weight
`=12.5xx10^(3)xx24=0.3g` of Mg
OR
`V_(HCl)=100cm^(3), V_(NaOH)=25cm^(3) , M_(HCl)=?, M_(NaOH)=0.1M`
and `V_(HCl)xxM_(HCl)=V_(NaOH)xxM_(NaOH)xxa_(NaOH)`
Where `a_(HCl)=` basicity of HCl `=1, aNaOH=` acidity of `NaOH=1`
`M_(HCl)=(25xx0.1)/100=0.025`m
Mass of HCl present in `100cm3=(M_(HCl)xx"Molecular wieight of " HCl)/10`
`=(0.025xx36.5)/10=0.0912g`