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 \({C_2}{H_4}\)
