Đáp án:
Bạn tham khảo lời giải ở dưới nhé !!!
Giải thích các bước giải:
\(\begin{array}{l}
8.\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
a)\\
{n_{{H_2}}} = 0,1mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,1mol \to {m_{Fe}} = 5,6g\\
\to {m_{Cu}} = 6,4g\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,2mol\\
\to C{M_{HCl}} = \dfrac{{0,2}}{{0,2}} = 1M\\
c)\\
\% {m_{Fe}} = \dfrac{{5,6}}{{12}} \times 100\% = 46,67\% \\
\% {m_{Cu}} = 100\% - 46,67\% = 53,33\% \\
9.\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = 0,045mol\\
a)\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,09mol\\
\to {m_{FeC{l_3}}} = 14,625g\\
b)\\
{n_{HCl}} = 6{n_{F{e_2}{O_3}}} = 0,27mol\\
\to {m_{HCl}} = 9,855g\\
\to C{\% _{{\rm{dd}}HCl}} = \dfrac{{9,855}}{{500}} \times 100\% = 1,971\% \\
c)\\
{m_{{\rm{dd}}}} = {m_{F{e_2}{O_3}}} + {m_{{\rm{dd}}HCl}} = 507,2g\\
\to C{\% _{FeC{l_3}}} = \dfrac{{14,625}}{{507,2}} \times 100\% = 2,88\% \\
10.\\
a)CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
b)\\
{n_{CuO}} = 0,1mol\\
{n_{{H_2}S{O_4}}} = {n_{CuO}} = 0,1mol\\
\to C{M_{{H_2}S{O_4}}} = \dfrac{{0,1}}{{0,2}} = 0,5M\\
c)\\
{n_{{\rm{CuS}}{O_4}}} = {n_{CuO}} = 0,1mol\\
\to {m_{{\rm{CuS}}{O_4}}} = 16g
\end{array}\)
