Đáp án:

 \(\begin{array}{l}
a)\\
{V_{{H_2}}} = 2,24l\\
b)\\
{m_{Zn}} = 6,5g\\
c)\\
{V_{{H_2}}} = 2,24l
\end{array}\)

Câu d em xem lại đề

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
ZnO + {H_2} \xrightarrow{t^0} Zn + {H_2}O\\
{n_{ZnO}} = \dfrac{{8,1}}{{81}} = 0,1\,mol\\
{n_{{H_2}}} = {n_{ZnO}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
b)\\
{n_{Zn}} = {n_{ZnO}} = 0,1\,mol\\
{m_{Zn}} = 0,1 \times 65 = 6,5g\\
c)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{HCl}} = \dfrac{{200 \times 7,3\% }}{{36,5}} = 0,4\,mol\\
{n_{Zn}} < \dfrac{{{n_{HCl}}}}{2} \Rightarrow  \text{ HCl dư }\\
{n_{{H_2}}} = {n_{Zn}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
d)\\
F{e_x}{O_y} + y{H_2} \xrightarrow{t^0}  xFe + y{H_2}O\\
{n_O} = {n_{{H_2}}} = 0,1\,mol\\
{m_{Fe}} = 3,24 - 0,1 \times 16 = 1,64g\\
{n_{Fe}} = \dfrac{{1,64}}{{56}} = 0,029\,mol
\end{array}\)