$$1.0 g$$  of magnesium is burnt with $$0.56 g$$  of oxygen in a closed vessel. Which reactant is left in excess and how much?
$$\left( {{\text{At}}{\text{. weight of}}\,Mg = 24,\,O = 16} \right)$$

Solution :
The balanced chemical equation is
\[\underset{\begin{smallmatrix} \\ 24\,g \end{smallmatrix}}{\mathop{Mg}}\,+\underset{16\,g}{\mathop{\frac{1}{2}{{O}_{2}}}}\,\to \underset{\begin{smallmatrix} \\ 40\,g \end{smallmatrix}}{\mathop{MgO}}\,\]
From the above equation, it is clear that,
$$24 g$$  of $$Mg$$  reacts with $$16 g$$  of $${{O_2}}$$.
Thus, $$1.0$$ $$g$$ of $$Mg$$  reacts with
$$\frac{{16}}{{24}}g\,\,{\text{of}}\,{O_2} = 0.67\,g\,{\text{of}}\,{O_2}.$$
But only $$0.56 g$$  of $${{O_2}}$$  is available which is less than $$0.67 g.$$  Thus $${O_2}$$  is the limiting reagent.
Further, $$16$$ $$g$$ of $${{O_2}}$$  reacts with $$24 g$$  of $$Mg.$$
∴ $$0.56 g$$  of $${{O_2}}$$  will react with
$$\eqalign{ & Mg = \frac{{24}}{{16}} \times 0.56 \cr & \,\,\,\,\,\,\,\,\,\, = 0.84\,g \cr} $$
∴ Amount of $$Mg$$  left unreacted
$$\eqalign{ & = \left( {1.0 - 0.84} \right)g\,\,Mg \cr & = 0.16\,g\,Mg \cr} $$