Đáp án:
1) mMgCl2=19 gam
V H2=4,48 lít
2) SO2
Giải thích các bước giải:
1) \[\begin{gathered}
Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2} \hfill \\
{n_{Mg}} = \frac{{4,8}}{{24}} = 0,2{\text{ mol}} \to {{\text{n}}_{Mg}} = {n_{MgC{l_2}}} = {n_{{H_2}}} = 0,2{\text{ mol}} \hfill \\
\to {{\text{m}}_{MgC{l_2}}} = 0,2(24 + 35,5.2) = 19{\text{gam}} \hfill \\
\to {{\text{V}}_{{H_2}}} = 0,2.22,4 = 4,48\;{\text{lit}} \hfill \\
\end{gathered} \]
2) Oxit có dạng
\[\begin{gathered}
{S_x}{O_y} \to {M_{{S_x}{O_y}}} = 32x + 16y = 64 \hfill \\
\to \% {m_S} = \frac{{32x}}{{64}} = 50\% \to x = 1 \to y = 2 \hfill \\
\to S{O_2} \hfill \\
\end{gathered} \]