Đáp án:
\(\begin{array}{l}
a)\\
{m_{Fe}} = 28g\\
{m_{Zn}} = 32,5g\\
b)\\
{V_{{H_2}}} = 22,4l
\end{array}\)
\(\begin{array}{l}
c)\\
{m_{FeC{l_2}}} = 63,5g\\
{m_{ZnC{l_2}}} = 68g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{m_{Fe}} = \% Fe \times {m_{hh}} = 46,289\% \times 60,5 = 28g\\
{m_{Zn}} = {m_{hh}} - {m_{Fe}} = 60,5 - 28 = 32,5g\\
b)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Zn}} = \dfrac{m}{M} = \dfrac{{32,5}}{{65}} = 0,5mol\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{28}}{{56}} = 0,5mol\\
{n_{{H_2}}} = {n_{Fe}} + {n_{Zn}} = 1mol\\
{V_{{H_2}}} = n \times 22,4 = 1 \times 22,4 = 22,4l
\end{array}\)
\(\begin{array}{l}
c)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,5mol\\
{m_{FeC{l_2}}} = n \times M = 0,5 \times 127 = 63,5g\\
{n_{ZnC{l_2}}} = {n_{Fe}} = 0,5mol\\
{m_{ZnC{l_2}}} = n \times M = 0,5 \times 136 = 68g
\end{array}\)
