\(n_{H_2SO_4}=\dfrac{200.9,8\%}{98}=0,2\left(mol\right)\)
\(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
0,15 < 0,2 ( mol )
0,15 0,15 0,15 ( mol )
`->` A gồm `CuSO_4` và `H_2SO_4`
\(m_{ddspứ}=200+12=212\left(g\right)\)
\(\left\{{}\begin{matrix}C\%_{CuSO_4}=\dfrac{0,15.160}{212}.100=11,32\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,2-0,15\right).98}{212}.100=2,31\%\end{matrix}\right.\)