Đáp án:
\(\begin{array}{l}
a)\\
{C_M}FeC{l_2} = 2,5M\\
b)\\
{V_{{H_2}}} = 11,2l\\
c)\\
{m_{{\rm{dd}}NaOH}} = 200g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = \dfrac{{28}}{{56}} = 0,5\,mol\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,5\,mol\\
{C_M}FeC{l_2} = \dfrac{{0,5}}{{0,2}} = 2,5M\\
b)\\
{n_{{H_2}}} = {n_{Fe}} = 0,5\,mol\\
{V_{{H_2}}} = 0,5 \times 22,4 = 11,2l\\
c)\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{HCl}} = 2{n_{Fe}} = 1\,mol\\
{n_{NaOH}} = {n_{HCl}} = 1\,mol\\
{m_{{\rm{dd}}NaOH}} = \dfrac{{1 \times 40}}{{20\% }} = 200g
\end{array}\)