Đáp án:
\(2/\\ C\%_{CuSO_4}=3,15\%\\ C\%_{H_2SO_4}(\text{dư})=16,36\%\)
Giải thích các bước giải:
\(1/\\ \text{Oxit axit: SO$_2$, CO$_2$, N$_2$O, N$_2$O$_5$, P$_2$O$_5$, SO$_3$}\\ \text{Oxit bazơ: K$_2$O, MgO, CaO, CuO, Fe$_2$O$_3$}\\ 2/\\ PTHH:CuO+H_2SO_4\to CuSO_4+H_2O\\ n_{CuO}=\dfrac{1,6}{80}=0,02\ mol.\\ n_{H_2SO_4}=\dfrac{100\times 20\%}{98}≈0,2\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,02}{1}<\dfrac{0,2}{1}\\ ⇒H_2SO_4\ dư.\\ Theo\ pt:\ n_{CuSO_4}=n_{CuO}=0,02\ mol.\\ m_{\text{dd spư}}=m_{CuO}+m_{\text{dd $H_2SO_4$}}=1,6+100=101,6\ g.\\ n_{H_2SO_4}(dư)=0,2-0,02=0,18\ mol.\\ ⇒C\%_{CuSO_4}=\dfrac{0,02\times 160}{101,6}\times 100\%=3,15\%\\ C\%_{H_2SO_4}(\text{dư})=\dfrac{0,18\times 98}{101,6}\times 100\%=16,36\%\)
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