Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 56,76\% \\
\% {m_{Cu}} = 43,24\% \\
b)\\
{m_{FeC{l_2}}} = 38,1g\\
c)\\
x = 2,4\\
d)\\
{C_M}FeC{l_2} = 0,12M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,3\,mol\\
{m_{Fe}} = 0,3 \times 56 = 16,8g\\
{m_{Cu}} = 29,6 - 16,8 = 12,8g\\
\% {m_{Fe}} = \dfrac{{16,8}}{{29,6}} \times 100\% = 56,76\% \\
\% {m_{Cu}} = 100 - 56,76 = 43,24\% \\
b)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,3\,mol\\
{m_{FeC{l_2}}} = 0,3 \times 127 = 38,1g\\
c)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6\,mol\\
{C_M}HCl = \dfrac{{0,6}}{{0,25}} = 2,4M \Rightarrow x = 2,4\\
d)\\
{C_M}FeC{l_2} = \dfrac{{0,3}}{{0,25}} = 0,12M
\end{array}\)