Đáp án:
\(\begin{array}{l}
11)B\\
12)\,D\\
13)A\\
14)D\\
15)C
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
11)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{56}}{{22,4}} = 2,5\,mol\\
{n_{HCl}} = 2{n_{{H_2}}} = 5\,mol\\
m = 83 + 5 \times 36,5 - 2,5 \times 2 = 260,5g\\
13)\\
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
{n_{KMn{O_4}}} = \dfrac{{1,58}}{{158}} = 0,01\,mol\\
{n_{C{l_2}}} = 0,01 \times \frac{5}{2} = 0,025\,mol\\
{V_{C{l_2}}} = 0,025 \times 22,4 = 0,56l\\
14)\\
{n_{C{l_2}}} = \dfrac{{0,336}}{{22,4}} = 0,015\,mol\\
{n_{{H_2}}} = \dfrac{{0,56}}{{22,4}} = 0,025\,mol\\
\text{ Lập tỉ lệ }:\dfrac{{{n_{C{l_2}}}}}{1} < \dfrac{{{n_{{H_2}}}}}{1}(0,015 < 0,025)\\
\Rightarrow \text{ $H_2$ dư tính theo $Cl_2$ } \\
C{l_2} + {H_2} \to 2HCl\\
{n_{HCl}} = 2{n_{C{l_2}}} = 0,03\,mol\\
H = \dfrac{{0,448}}{{0,03 \times 22,4}} \times 100\% = 66,67\% \\
15)\\
2X + nC{l_2} \to 2XC{l_n}\\
{n_{C{l_2}}} = \dfrac{{0,448}}{{22,4}} = 0,02\,mol\\
{n_{XC{l_n}}} = 0,02 \times \dfrac{2}{n} = \dfrac{{0,04}}{n}\,mol\\
{M_{XC{l_n}}} = \dfrac{{1,9}}{{\dfrac{{0,04}}{n}}} = 47,5n\,g/mol\\
\Rightarrow {M_X} + 35,5n = 47,5n \Rightarrow {M_X} = 12n\,g/mol\\
n = 2 \Rightarrow {M_X} = 24g/mol \Rightarrow X:Mg
\end{array}\)
