Answer:
The amount of Mg was enough
Explanation:
In this case, we have to start with the reaction between [tex]Mg[/tex] and [tex]CuO[/tex], so:
[tex]CuO~+~Mg~->~MgO~+~Cu[/tex]
If we check the reaction is already balanced. Now, we can do some stoichiometry to calculate the amount of Mg. The first step is the number of moles of [tex]CuO[/tex]. To this we have to calculate the molar mass of [tex]CuO[/tex] first, so:
Cu: 63.55 g/mol and O: 16 g/mol. So, (63.55+16)= 79.55 g/mol.
Now, we can calculate the moles:
[tex]1.525~g~CuO\frac{1~mol~CuO}{79.55~g~CuO}=0.0192~mol~CuO[/tex]
The molar ratio between [tex]Mg[/tex] and [tex]CuO[/tex] is 1:1, so:
[tex]0.0192~mol~CuO=0.0192~mol~Mg[/tex].
Now we can calculate the mass of Mg if we know the atomic mass of Mg (24.305 g/mol). So:
[tex]0.0192~mol~Mg\frac{24.305~g~Mg}{1~mol~Mg}=0.466~g~Mg[/tex]
With this in mind, the student added enough Mg to recover all the Cu.
Note: The HCl doesn't take a role in the reaction. The function of HCl is to dissolve the [tex]CuO[/tex].
I hope it helps!