a)
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\); \(n_{H_2SO_4}=\dfrac{200.10\%}{98}=\dfrac{10}{49}\left(mol\right)\)
PTHH: CuO + H2SO4 --> CuSO4 + H2O
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{\dfrac{10}{49}}{1}\) => CuO hết, H2SO4 dư
Theo PTHH: \(n_{H_2SO_4\left(pư\right)}=n_{CuO}=0,1\left(mol\right)\)
=> \(m_{H_2SO_4\left(dư\right)}=\left(\dfrac{10}{49}-0,1\right).98=10,2\left(g\right)\)
b) mdd sau pư = 8 + 200 = 208 (g)
\(n_{CuSO_4}=n_{CuO}=0,1\left(mol\right)\Rightarrow m_{CuSO_4}=0,1.160=16\left(g\right)\)
\(\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{10,2}{208}.100\%=4,9\%\\C\%_{CuSO_4}=\dfrac{16}{208}.100\%=7,7\%\end{matrix}\right.\)
