describe an activity to show how the following metals can be arranged in the decreasing order of reactivity with dilute sulphuric acid: AL,ZN,CU,FE,MG

Dear User,
The more the reactive an element will be, more easily it will replace H from  H₂SO₄ and more will be its reactivity. By treating the metals with dilute solution of sulphuric acid, their order of reactivity can be arranged.
$$\begin{gathered} 2\mathrm{Al} + 3{\mathrm{H}}_2{\mathrm{SO}}_4 \to {\mathrm{Al}}_2({\mathrm{SO}}_4{)}_3 + 3{\mathrm{H}}_2 \\ \mathrm{Zn} + {\mathrm{H}}_2{\mathrm{SO}}_4 \to {\mathrm{ZnSO}}_4 + {\mathrm{H}}_2 \\ \mathrm{Cu} + {\mathrm{H}}_2{\mathrm{SO}}_4 \to \mathrm{No} \mathrm{reaction} \\ \mathrm{Fe} + {\mathrm{H}}_2{\mathrm{SO}}_{4 }\to {\mathrm{FeSO}}_{4 }+ {\mathrm{H}}_2 \\ \mathrm{Mg} + {\mathrm{H}}_2{\mathrm{SO}}_4 \to {\mathrm{MgSO}}_{4 }+ {\mathrm{H}}_2 \end{gathered}$$
Of all the metals, Cu do not reacts will sulphuric acid, thus it is least reactive.
Now, the oder with which rest of the metals react with dilute sulphuric acid is :
Mg >Al >Zn > Fe
Therefore, decreasing order of reactivity of metals is :
Mg >Al >Zn > Fe>Cu