Đáp án:
Câu 4: $B$
Câu 5: $C$
Giải thích các bước giải:
Câu 4: $B$
PTHH: $NaOH + MgC{l_2} \to NaCl + Mg{(OH)_2} \downarrow $
Câu 5: $C$
${n_{{H_2}}} = 0,17\,\,mol$
Mg phản ứng với $HCl$ sinh ra khí $H_2$
PTHH:
$\begin{gathered} Mg + 2HCl \to MgC{l_2} + {H_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ 0,17 \leftarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,17\,\,\,\,\,\,\,\,mol \hfill \\ \end{gathered} $
$\begin{gathered} {n_{Mg}} = 0,17\,\,mol \to {m_{Mg}} = 0,17.24 = 4,08\,\,g \hfill \\ \to {m_{BaO}} = 20 - 4,08 = 15,92\,\,g\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \to \% {m_{BaO}} = \frac{{15,92}}{{20}}.100\% = 79,6\% \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} $
Chọn $C$