a) Ta có: 

\({n_{Cu}} = \frac{{12,8}}{{64}} = 0,2{\text{ mol; }}{{\text{n}}_S} = \frac{{6,4}}{{32}} = 0,2{\text{ mol; }}{{\text{n}}_O} = \frac{{12,8}}{{16}} = 0,8{\text{ mol}}\)

\( \to {n_{Cu}}:{n_S}:{n_O} = 0,2:0,2:0,8 = 1:1:4\)

Hợp chất có dạng 

\({(CuS{O_4})_n} \to 160n = 160 \to n = 1 \to CuS{O_4}\)

b) \(B + {O_2}\xrightarrow{{}}C{O_2} + {H_2}O\)

Ta có: \({M_B} = 14.2 = 28\)

Ta có: 

\({n_{C{O_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol = }}{{\text{n}}_C}{\text{; }}{{\text{n}}_{{H_2}O}} = \frac{{3,6}}{{18}} = 0,2{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 0,4{\text{ mol}} \to {{\text{n}}_C}:{n_H} = 0,2:0,4 = 1:2\)

A có dạng \({C_n}{H_{2n}}{O_z} \to 14n + 16z = 28\)

Nếu z=1 suy ra n lẻ 

Nếu z=0 suy ra n=2.

Vậy B là \({C_2}{H_4}\)

\(\begin{array}{l}
a)\\
CTHH:Cu_xS_yO_z\\
x:y:z=\frac{12,8}{64}:\frac{6,4}{32}:\frac{12,8}{16}=0,2:0,2:0,8=1:1:4\\
\to CTN:(CuSO_4)_n\\
M=160\to (63+32+16.4).n=160\\
\to n=1\to CTHH:CuSO_4\\
b)\\
n_C=n_{CO_2}=\frac{4,48}{22,4}=0,2(mol)\\
n_H=2n_{H_2O}=2.\frac{3,6}{18}=0,4(mol)\\
TH_1:\,A\,gom\,C,H,O\\
\to A:C_xH_yO_z\\
\to x:y=0,2:0,4=1:2\\
\to A:C_nH_{2n}O_z\\
\to 12n+2n+16z=14.2=28\\
\to 14n+16z=28(\rm vô\,nghiệm\,nguyên\,dương)\\
TH_2:\,A\,gom\,C,H\\
\to n_C:n_H=0,2:0,4=1:2\\
\to CTN:(CH_2)_n\\
M=14.2=28\\
\to (12+2).n=28\\
\to n=2\\
\to CTHH:C_2H_4
\end{array}\)