Đáp án:
\(\begin{array}{l}
b)\\
{V_{{H_2}}} = 1,12l\\
c)\\
{m_{Fe}} = 2,8g\\
{m_{FeC{l_2}}} = 6,35g\\
{m_{{H_2}}} = 0,1g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{n_{Fe}} = \dfrac{{5,6}}{{56}} = 0,1\,mol\\
{n_{HCl}} = 0,2 \times 0,5 = 0,1\,mol\\
{n_{Fe}} > \dfrac{{{n_{HCl}}}}{2} \Rightarrow Fe \text{ dư }\\
{n_{{H_2}}} = \dfrac{{{n_{HCl}}}}{2} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{V_{{H_2}}} = 0,05 \times 22,4 = 1,12l\\
c)\\
{n_{Fe}} \text{ dư }= 0,1 - \dfrac{{0,1}}{2} = 0,05\,mol\\
{n_{FeC{l_2}}} = {n_{{H_2}}} = 0,05\,mol\\
{m_{Fe}} \text{ dư } = 0,05 \times 56 = 2,8g\\
{m_{FeC{l_2}}} = 0,05 \times 127 = 6,35g\\
{m_{{H_2}}} = 0,05 \times 2 = 0,1g
\end{array}\)
