Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

9,

\(\begin{array}{l}
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{HCl}} = \dfrac{{146 \times 5}}{{100 \times 36,5}} = 0,2mol\\
 \to {n_{CaC{O_3}}} = \dfrac{1}{2}{n_{HCl}} = 0,1mol\\
 \to {m_{CaC{O_3}}} = 10g\\
{n_{C{O_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,1mol\\
 \to {V_{C{O_2}}} = 2,24l\\
Ca{(OH)_2} + 2HCl \to CaC{l_2} + 2{H_2}O\\
{n_{Ca{{(OH)}_2}}} = \dfrac{{185 \times 10}}{{100 \times 74}} = 0,25mol\\
 \to {n_{Ca{{(OH)}_2}}} > \dfrac{{{n_{HCl}}}}{2}
\end{array}\)

Suy ra \(Ca{(OH)_2}\) dư 

\(\begin{array}{l}
 \to {n_{Ca{{(OH)}_2}}}dư= 0,25 - \dfrac{1}{2}{n_{HCl}} = 0,15mol\\
{n_{CaC{l_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,1mol\\
{m_{{\rm{dd}}}} = 185 + 146 = 331g\\
 \to C{\% _{Ca{{(OH)}_2}}}dư= \dfrac{{0,15 \times 74}}{{331}} \times 100\%  = 3,35\% \\
 \to C{\% _{CaC{l_2}}} = \dfrac{{0,1 \times 111}}{{331}} \times 100\%  = 3,35\% 
\end{array}\)

10,

\(\begin{array}{l}
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{{H_2}}} = 0,1mol\\
 \to {n_{Zn}} = {n_{{H_2}}} = 0,1mol\\
 \to {m_{Zn}} = 6,5g\\
 \to {m_{Cu}} = 10,5 - {m_{Zn}} = 10,5 - 6,5 = 4g\\
Zn + 2AgN{O_3} \to Zn{(N{O_3})_2} + 2Ag\\
Cu + 2AgN{O_3} \to Cu{(N{O_3})_2} + 2Ag\\
{n_{Cu}} = 0,0625mol\\
 \to {n_{Ag}} = 2 \times ({n_{Zn}} + {n_{Cu}}) = 0,325mol\\
 \to {m_{Ag}} = 35,1g
\end{array}\)

11,

Phần 1: 

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = 0,45mol\\
 \to \dfrac{3}{2}{n_{Al}} + {n_{Fe}} = {n_{{H_2}}} = 0,45mol
\end{array}\)

Phần 2: 

\(\begin{array}{l}
2Al + 2NaOH + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}\\
{m_{Fe}} = 8,4g\\
 \to {n_{Fe}} = 0,15mol\\
 \to {n_{Al}} = 0,2mol\\
 \to \% {m_{Al}} = \dfrac{{0,2 \times 2 \times 27}}{{27,6}} \times 100\%  = 39,13\% \\
 \to \% {m_{Cu}} = 100\%  - 39,13\%  = 60,87\% 
\end{array}\)