a, Ta có: \(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)
\(m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
Xét tỉ lệ: \(\dfrac{0,02}{1}< \dfrac{0,2}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{CuSO_4}=n_{H_2SO_4\left(pư\right)}=n_{CuO}=0,02\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,18\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CuSO_4}=0,02.160=3,2\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,18.98=17,64\left(g\right)\end{matrix}\right.\)
b, Ta có: m dd sau pư = 1,6 + 200 = 201,6 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{CuSO_4}=\dfrac{3,2}{201,6}.100\%\approx1,59\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{17,64}{201,6}.100\%=8,75\%\end{matrix}\right.\)
Bạn tham khảo nhé!
