Đáp án:
4 A
5 A
Giải thích các bước giải:
\(\begin{array}{l}
4)\\
Mg \to M{g^{ + 2}} + 2e\\
Al \to A{l^{ + 3}} + 3e\\
Zn \to Z{n^{ + 2}} + 2e\\
{S^{ + 6}} + 2e \to {S^{ + 4}}\\
C{l_2} + 2e \to 2C{l^ - }\\
{n_{S{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
BTe:3{n_{Al}} + 2{n_{Mg}} + 2{n_{Zn}} = 2{n_{S{O_2}}}(1)\\
BT\,e:3{n_{Al}} + 2{n_{Mg}} + 2{n_{Zn}} = 2{n_{C{l_2}}}(2)\\
(1),(2) \Leftrightarrow 2{n_{S{O_2}}} = 2{n_{C{l_2}}} \Leftrightarrow {n_{C{l_2}}} = {n_{S{O_2}}} = 0,2\,mol\\
BTKL:\\
{m_{hh}} + {m_{C{l_2}}} = {m_m} \Leftrightarrow {m_{hh}} = 19,2 - 0,2 \times 71 = 5g\\
m = 5 \times 2 = 10g\\
5)\\
M \to {M^{ + n}} + ne\\
2{N^{ + 5}} + 8e \to {N_2}O\\
{N^{ + 5}} + 3e \to NO\\
BT\,e:{n_M} \times n = 8{n_{{N_2}O}} + 3{n_{NO}}\\
\Leftrightarrow {n_M} = \dfrac{{5,1}}{n}\,mol\\
{M_M} = \dfrac{{45,9}}{{\dfrac{{5,1}}{n}}} = 9n\,g/mol\\
n = 3 \Rightarrow {M_M} = 27g/mol \Rightarrow M:Al
\end{array}\)
