Đáp án:
32 C
33 B
Giải thích các bước giải:
\(\begin{array}{l}
32)\\
R + 2HCl \to RC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_R} = {n_{{H_2}}} = 0,15\,mol\\
{M_R} = \dfrac{{4,4}}{{0,15}} = 29,333(g/mol)\\
\Rightarrow hh:Magie(Mg),Can\,xi(Ca)\\
33)\\
2R + 2{H_2}O \to 2ROH + {H_2}\\
ROH + HCl \to RCl + {H_2}O\\
{n_{{H_2}}} = \dfrac{{0,336}}{{22,4}} = 0,015\,mol\\
\Rightarrow {n_{ROH}} = 2{n_{{H_2}}} = 0,015 \times 2 = 0,03\,mol\\
{n_{RCl}} = {n_{ROH}} = 0,03\,mol\\
{M_{RCl}} = \dfrac{{2,075}}{{0,03}} = 69,16(g/mol)\\
\Rightarrow {M_R} = 69,16 - 35,5 = 33,66(g/mol)\\
\Rightarrow hh:Natri(Na),Kali(K)
\end{array}\)
