Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

\(\begin{array}{l}
1,\\
a,Ca\\
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
b,Zn,Ca,Fe\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Ca + 2HCl \to CaC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
c,Zn,Ca\\
Zn + FeS{O_4} \to ZnS{O_4} + Fe\\
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
Ca{(OH)_2} + FeS{O_4} \to CaS{O_4} + Fe{(OH)_2}\\
2,\\
Fe \to F{e^{3 + }} + 3e\\
{N^{5 + }} + 3e \to {N^{2 + }}\\
{n_{Fe}} = 0,1mol\\
 \to 3{n_{Fe}} = 3{n_{NO}}\\
 \to {n_{NO}} = {n_{Fe}} = 0,1mol\\
 \to {V_{NO}} = 2,24l\\
3,\\
R + 2{H_2}O \to R{(OH)_2} + {H_2}\\
{n_{{H_2}}} = 0,45mol\\
 \to {n_R} = {n_{{H_2}}} = 0,45mol\\
 \to {M_R} = 40\\
 \to Ca
\end{array}\)

4,

\(\begin{array}{l}
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = 0,65mol\\
 \to {n_{Mg}} + 1,5{n_{Al}} = {n_{{H_2}}} = 0,65mol\\
\left\{ \begin{array}{l}
24{n_{Mg}} + 27{n_{Al}} = 12,9\\
{n_{Mg}} + 1,5{n_{Al}} = 0,65
\end{array} \right.\\
 \to {n_{Mg}} = 0,2mol \to {n_{Al}} = 0,3mol\\
 \to {m_{Mg}} = 4,8g\\
 \to {m_{Al}} = 8,1g\\
 \to \% {m_{Mg}} = \dfrac{{4,8}}{{12,9}} \times 100\%  = 37,21\% \\
 \to \% {m_{Al}} = 100\%  - 37,21\%  = 62,79\% \\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,65mol\\
 \to {m_{{H_2}S{O_4}}} = 63,7g
\end{array}\)

5,

\(\begin{array}{l}
2Al + 3CuC{l_2} \to 2AlC{l_3} + 3Cu\\
{n_{CuC{l_2}}} = 0,15mol\\
 \to {n_{Al}}phản ứng= \dfrac{2}{3}{n_{CuC{l_2}}} = 0,1mol\\
 \to {n_{Cu}} = {n_{CuC{l_2}}} = 0,15mol\\
 \to {m_{tăng}} = {m_{Cu}} - {m_{Al}}phản ứng\\
 \to {m_{tăng}} = 0,15 \times 64 - 0,1 \times 27 = 6,9g
\end{array}\)