Em tham khảo nha :
\(\begin{array}{l}
a)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
ZnO + {H_2}S{O_4} \to ZnS{O_4} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,2mol\\
{m_{Mg}} = 0,2 \times 24 = 4,8g\\
\% Mg = \dfrac{{4,8}}{{25,8}} \times 100\% = 18,6\% \\
\% ZnO = 100 - 18,6 = 81,4\% \\
b)\\
{n_{ZnO}} = \dfrac{{25,8 - 4,8}}{{81}} = 0,26mol\\
{n_{{H_2}S{O_4}}} = {n_{Mg}} + {n_{ZnO}} = 0,2 + 0,26 = 0,46mol\\
{m_{{H_2}S{O_4}}} = 0,46 \times 98 = 45,08\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{45,08}}{{160}} \times 100\% = 28,175\% \\
c)\\
{m_{ddspu}} = 25,8 + 160 - 0,2 \times 2 = 185,4g\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,2mol\\
{m_{MgS{O_4}}} = 0,2 \times 120 = 24g\\
{n_{ZnS{O_4}}} = {n_{ZnO}} = 0,26mol\\
{m_{ZnS{O_4}}} = 0,26 \times 161 = 41,86g\\
C{\% _{MgS{O_4}}} = \dfrac{{24}}{{185,4}} \times 100\% = 12,9\% \\
C{\% _{ZnS{O_4}}} = \dfrac{{41,86}}{{185,4}} \times 100\% = 22,6\%
\end{array}\)
