Đáp án:
\(\begin{array}{l}
b)\\
{V_{{H_2}}} = 4,48l\\
c)\\
{m_{Fe}} = 7,47g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{m}{M} = \dfrac{{13}}{{65}} = 0,2mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
{V_{{H_2}}} = n \times 22,4 = 2 \times 22,4 = 4,48l\\
c)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{Fe}} = \dfrac{2}{3}{n_{{H_2}}} = \dfrac{2}{{15}}mol\\
{m_{Fe}} = n \times M = \frac{2}{{15}} \times 56 = 7,47g
\end{array}\)