Đáp án:

\(\begin{array}{l}
46,D\\
47,A\\
48,D\\
49,C\\
50,D\\
51,A\\
52,D\\
53,A\\
54,B\\
55,B
\end{array}\)

Giải thích các bước giải:

46,

\(\begin{array}{l}
Ca{(OH)_2} + 2HCl \to CaC{l_2} + 2{H_2}O\\
Ca{(OH)_2} + C{O_2} \to CaC{{\rm{O}}_3} + {H_2}O
\end{array}\)

47,

\(\begin{array}{l}
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,05mol\\
 \to {n_{MgO}} + 3{n_{A{l_2}{O_3}}} = 0,05
\end{array}\)

Mặt khác, ta có: \(40{n_{MgO}} + 102{n_{A{l_2}{O_3}}} = 1,82\)

Giải hệ phương trình ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{n_{MgO}} + 3{n_{A{l_2}{O_3}}} = 0,05\\
40{n_{MgO}} + 102{n_{A{l_2}{O_3}}} = 1,82
\end{array} \right.\\
 \to {n_{MgO}} = 0,02mol \to {n_{A{l_2}{O_3}}} = 0,01mol\\
 \to \% {m_{MgO}} = \dfrac{{0,02 \times 40}}{{1,82}} \times 100\%  = 43,96\% \\
 \to \% {m_{A{l_2}{O_3}}} = 100\%  - 43,96\%  = 56,04\% 
\end{array}\)

49,

\(\begin{array}{l}
{n_{C{O_2}}} = 0,2mol\\
{n_{NaOH}} = 0,2mol\\
 \to \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = 1\\
C{O_2} + NaOH \to NaHC{O_3}
\end{array}\)

50,

\(\begin{array}{l}
Zn + {H_2}S{O_4} \to Zn{\rm{S}}{O_4} + {H_2}\\
CuO + {H_2} \to Cu + {H_2}O
\end{array}\)

51,

\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,1mol\\
 \to {n_{HCl}} = 2{n_{{H_2}}} = 0,2mol\\
 \to C{M_{HCl}} = 1M
\end{array}\)

53,

\(\begin{array}{l}
K{\rm{O}}H + HCl \to KCl + {H_2}O\\
2K{\rm{O}}H + CuC{l_2} \to 2KCl + Cu{(OH)_2}\\
K{\rm{O}}H + Al{(OH)_3} \to KAl{O_2} + 2{H_2}O
\end{array}\)

54,

\(\begin{array}{l}
N{a_2}S{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + S{O_2} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = 0,1mol\\
 \to {n_{S{O_2}}} = {n_{N{a_2}S{O_3}}} = 0,1mol\\
 \to {V_{S{O_2}}} = 2,24l
\end{array}\)

55,

\(\begin{array}{l}
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
NaOH + HCl \to NaCl + {H_2}O
\end{array}\)