\(Mg +2 HCl \rightarrow MgCl _{2}+ H _{2}\)
Mole of magnesium \(=0.125 mol\)
Mole of \(HCl\)
\(=\left(\frac{25}{1000}\right) dm ^{3} \times 4.0 moldm ^{-3}\)
Since \(1 mol\) of \(Mg\) is required for \(2 mol\) of \(HCl\) (according to the equation), \(0.125 mol\) of \(Mg\) will require \((0.125 \times 2) mol =0.25 mol\) 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, \(2 mol\) of \(HC /\) produce 1 mol of hydrogen gas.
The amount of hydrogen that will be produced from \(0.1\) mole of \(HCl\)
$$
=\left(\frac{0.1}{2}\right) mol =0.05 mol
$$