a)
\(n_{CuO}=\dfrac{0,8}{80}=0,01\left(mol\right)\\ m_{H_2SO_4}=50.10\%=5\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{5}{98}\left(mol\right)\)
PTHH:
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
ban đầu: 0,01 \(\dfrac{5}{98}\)
sau pư 0 \(\dfrac{201}{4900}\) 0,01
b)
\(m_{dd.spu}=0,8+50=50,8\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4.dư}=\dfrac{\dfrac{201}{4900}.98}{50,8}.100\%=7,91\%\\C\%_{CuSO_4}=\dfrac{0,01.160}{50,8}.100\%=3,15\%\end{matrix}\right.\)