Em tham khảo nha:
\(\begin{array}{l}
{M_{AO}} = \dfrac{2}{3} \times {M_{MgS{O_4}}} = \dfrac{2}{3} \times 120 = 80(g/mol)\\
\Rightarrow {M_A} = 80 - 16 = 64(g/mol)\\
\Rightarrow A:Cu\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
B:Cu{(OH)_2}\\
C:S{O_2}\\
CuS{O_4} + Ba{(OH)_2} \to BaS{O_4} + Cu{(OH)_2}\\
D:Cu{(OH)_2}\\
Cu{(OH)_2} \xrightarrow{t^0} CuO + {H_2}O\\
E:CuO\\
S{O_2} + NaOH \to NaHS{O_3}\\
F:NaHS{O_3}\\
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
L:N{a_2}S{O_3}\\
2S{O_2} + {O_2} \xrightarrow{t^0,V_2O_5} 2S{O_3}\\
M:S{O_3}
\end{array}\)
