Em tham khảo nha :

\(\begin{array}{l}
a)\\
Mg + ZnS{O_4} \to MgS{O_4} + Zn\\
MgS{O_4} + 2NaOH \to N{a_2}S{O_4} + Mg{(OH)_2}\\
2Al + 3ZnS{O_4} \to A{l_2}{(S{O_4})_3} + 3Zn(1)\\
A{l_2}{(S{O_4})_3} + 6NaOH \to 2Al{(OH)_3} + 3N{a_2}S{O_4}\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O\\
{n_{Mg{{(OH)}_2}}} = \dfrac{{0,87}}{{58}} = 0,015mol\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{1,008}}{{22,4}} = 0,045mol\\
{n_{Zn}} = {n_{{H_2}}} = 0,045mol\\
{n_{Zn(1)}} = 0,045 - 0,015 = 0,03mol\\
{n_{Al}} = \dfrac{2}{3}{n_{Zn}} = 0,02mol\\
{m_{Al}} = 0,02 \times 27 = 0,54g\\
{m_{Mg}} = 0,015 \times 24 = 0,36g\\
b)\\
{n_{ZnS{O_4}}} = 0,3 \times 0,2 = 0,06mol\\
{n_{ZnS{O_4}d}} = 0,06 - 0,03 - 0,015 = 0,015mol\\
ZnS{O_4} + 2NaOH \to Zn{(OH)_2} + N{a_2}S{O_4}\\
{n_{NaOH}} = 2{n_{MgS{O_4}}} + 6{n_{A{l_2}{{(S{O_4})}_3}}} + 2{n_{ZnS{O_4}d}}\\
 \Rightarrow {n_{NaOH}} = 2{n_{Mg}} + 3{n_{Al}} + 2{n_{ZnS{O_4}d}} = 0,12mol
\end{array}\)