Đáp án:
\(\begin{array}{l}
7)\\
{m_{Fe}} = 22,4g\\
{m_{FeC{l_2}}} = 50,8g\\
8)\\
a)\\
{V_{{O_2}}} = 4,48l\\
b)\\
{m_{{H_2}S{O_4}}} = 39,2g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
7)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{CuO}} = \dfrac{{32}}{{80}} = 0,4\,mol \Rightarrow {n_{{H_2}}} = 0,4\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,4\,mol\\
{m_{Fe}} = 0,4 \times 56 = 22,4g\\
{n_{FeC{l_2}}} = {n_{{H_2}}} = 0,4\,mol\\
{m_{FeC{l_2}}} = 0,4 \times 127 = 50,8g\\
8)\\
a)\\
3Fe + 2{O_2} \to F{e_3}{O_4}\\
{n_{Fe}} = \dfrac{{16,8}}{{56}} = 0,3\,mol\\
\Rightarrow {n_{{O_2}}} = \frac{{0,3 \times 2}}{3} = 0,2mol\\
{V_{{O_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
F{e_3}{O_4} + 4{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + FeS{O_4} + 4{H_2}O\\
{n_{F{e_3}{O_4}}} = \dfrac{{0,3}}{3} = 0,1\,mol\\
\Rightarrow {n_{{H_2}S{O_4}}} = 0,1 \times 4 = 0,4\,mol\\
{m_{{H_2}S{O_4}}} = 0,4 \times 98 = 39,2g
\end{array}\)
