$Mg+2HCl\to MgC{l}_2+H_2$
Mole of magnesium $=0.125mol$
Mole of $HCl$
$=\left(\frac{25}{1000}\right)d{m}^3\times 4.0mold{m}^{-3}$
Since $1mol$ of $Mg$ is required for $2mol$ of $HCl$ (according to the equation), $0.125mol$ of $Mg$ will require $(0.125\times 2)mol=0.25mol$ of $HCl$
Since the amount of $HCl$ available is not up to this, $HCl$ is the limiting reagent. A limiting reagent determines the amount of products formed.gain, according to the equation, $2mol$ of $HC/$ produce 1 mol of hydrogen gas.
The amount of hydrogen that will be produced from $0.1$ mole of $HCl$
$$=\left(\frac{0.1}{2}\right)mol=0.05mol$$