Em tham khảo nha :

\(\begin{array}{l}
2)\\
{n_{CO}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_{C{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{M_A} = \dfrac{{0,05 \times 28 + 0,15 \times 44}}{{0,05 = 0,15}} = 40dvC\\
{d_{A/{H_2}}} = \dfrac{{{M_A}}}{{{M_{{H_2}}}}} = \dfrac{{40}}{2} = 20\\
b)\\
hh:CO(a\,mol),C{O_2}(b\,mol)\\
{M_A} = 20{M_{{H_2}}} = 40dvC\\
 \Rightarrow \dfrac{{28a + 44b}}{{a + b}} = 40\\
 \Rightarrow \dfrac{a}{b} = \dfrac{{44 - 40}}{{40 - 28}} = \dfrac{1}{3}\\
 \Rightarrow {V_{CO}}:{V_{C{O_2}}} = 1:3\\
3)\\
CuO + CO \to Cu + C{O_2}\\
FeO + CO \to Fe + C{O_2}\\
F{e_2}{O_3} + 3CO \to 2Fe + 3C{O_2}\\
F{e_3}{O_4} + 4CO \to 3Fe + 4C{O_2}\\
{n_{C{O_2}}} = \dfrac{{13,2}}{{44}} = 0,3mol\\
{n_{CO}} = {n_{C{O_2}}} = 0,3mol\\
{m_{CO}} = 0,3 \times 28 = 8,4g\\
\text{Theo định luật bảo toàn khối lượng ta có :}\\
{m_X} + {m_{CO}} = {m_Y} + {m_{C{O_2}}}\\
 \Rightarrow {m_X} = 40 + 13,2 - 8,4 = 44,8g
\end{array}\)

Câu còn lại làm sau cho , bh mắc r