Đáp án:
\(\begin{array}{l}
b)\\
{V_{{H_2}}} = 1,792l\\
c)\\
{m_{FeC{l_2}}} = 10,16g\\
d)\\
{C_\% }FeC{l_2} = 4,97\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{m_{HCl}} = 200 \times 3,65\% = 7,3g\\
{n_{HCl}} = \dfrac{{7,3}}{{36,5}} = 0,2\,mol\\
H = 80\% \Rightarrow {n_{{H_2}}} = \dfrac{{0,2}}{2} \times 80\% = 0,08\,mol\\
{V_{{H_2}}} = 0,08 \times 22,4 = 1,792l\\
c)\\
{n_{FeC{l_2}}} = {n_{{H_2}}} = 0,08\,mol\\
{m_{FeC{l_2}}} = 0,08 \times 127 = 10,16g\\
d)\\
{n_{Fe}} = {n_{FeC{l_2}}} = 0,08\,mol\\
{m_{ddspu}} = 0,08 \times 56 + 200 - 0,08 \times 2 = 204,32g\\
{C_\% }FeC{l_2} = \dfrac{{10,16}}{{204,32}} \times 100\% = 4,97\%
\end{array}\)