Em tham khảo nha :

$ZnO$ chứ không phải $Zn$ nha em

\(\begin{array}{l}
a)\\
ZnO + {H_2}S{O_4} \to ZnS{O_4} + {H_2}O\\
b)\\
{n_{ZnO}} = \dfrac{{16,2}}{{81}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = {n_{ZnO}} = 0,2mol\\
{m_{{H_2}S{O_4}}} = 0,2 \times 98 = 19,6g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{19,6}}{{200}} \times 100\%  = 9,8g\\
c)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{m_{{H_2}S{O_4}}} = \dfrac{{100 \times 9,8}}{{100}} = 9,8g\\
{n_{{H_2}S{O_4}}} = \frac{{9,8}}{{98}} = 0,1mol\\
{n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,2mol\\
{C_{{M_{NaOH}}}} = \dfrac{{0,2}}{{0,2}} = 1l
\end{array}\)

Xin phép cho `m_{Zn}` lên `16,25` cho tròn ạ 

`n_{Zn}=\frac{16,25}{65}=0,25(mol)`

`Zn+H_2SO_4->ZnSO_4+H_2`

`b,`

Theo phương trình

`n_{H_2SO_4}=n_{Zn}=0,25(mol)`

`->C%_{H_2SO_4}=\frac{0,25.98}{200}.100=12,25%`

`c,`

`n_{H_2SO_4}=\frac{100.12,25%}{98}=0,125(mol)`

`H_2SO_4+2NaOH->Na_2SO_4+2H_2O`

Theo phương trình

`n_{NaOH}=2n_{H_2SO_4}=0,125.2=0,25(mol)`

`->V_{NaOH}=\frac{0,25}{0,2}=1,25(l)`