`n_(CuO)=\frac{8}{80}=0,1(mol)`

`CuO+2HCl->CuCl_2+H_2O`

`0,1`        `0,2`       `0,1`

`m_(HCl)=0,2.36,5=7,3(g)`

`C%_(HCl)=\frac{7,3}{100}.100=7,3%`

`m_(CuCl_2)=0,1.135=13,5(g)`

`B2,`

`n_(H_2SO_4)=\frac{19,6}{98}=0,2(mol)`

`MO+H_2SO_4->MSO_4+H_2O`

`0,2`         `0,2`

`M_(MO)=\frac{8}{0,2}=40`

`=>M+16=40`

`=>M=24`

Vậy `M` là `Mg`

1)

Phản ứng xảy ra:

\(CuO + 2HCl\xrightarrow{{}}CuC{l_2} + {H_2}O\)

Ta có: 

\({n_{CuO}} = \frac{8}{{64 + 16}} = 0,1{\text{ mol}}\)

\( \to {n_{CuC{l_2}}} = {n_{Cu}} = 0,1{\text{ mol}}\)

\( \to {m_{CuC{l_2}}} = 0,1.(64 + 35,5.2) = 13,5{\text{ gam}}\)

\({n_{HCl}} = 2{n_{CuO}} = 0,2{\text{ mol}}\)

\( \to {m_{HCl}} = 0,2.36,5 = 7,3{\text{ gam}}\)

\( \to C{\% _{HCl}} = \frac{{7,3}}{{100}} = 7,3\% \)

2)

Phản ứng xảy ra:

\(MO + {H_2}S{O_4}\xrightarrow{{}}MS{O_4} + {H_2}O\)

Ta có: 

\({n_{{H_2}S{O_4}}} = \frac{{19,6}}{{98}} = 0,2{\text{ mol = }}{{\text{n}}_{MO}}\)

\( \to {M_{MO}} = \frac{8}{{0,2}} = 40= {M_M} + 16\)

\( \to {M_M} = 24 \to M:Mg\)

Vậy oxit là \(MgO\)