Đáp án:
\(\begin{array}{l} a,\ m_{CuSO_4}=24\ g.\\ b,\ C\%_{CuSO_4}=26,55\%\\ C\%_{H_2SO_4\text{(dư)}}=5,42\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\\ PTHH:CuO+H_2SO_4\to CuSO_4+H_2O\\ n_{CuO}=\dfrac{12}{80}=0,15\ mol.\\ n_{H_2SO_4}=\dfrac{78,4\times 25\%}{98}=0,2\ mol.\\ \text{Lập tỉ lệ}\ n_{CuO}:n_{H_2SO_4}=\dfrac{0,15}{1}<\dfrac{0,2}{1}\\ \Rightarrow H_2SO_4\ \text{dư.}\\ Theo\ pt:\ n_{CuSO_4}=n_{CuO}=0,15\ mol.\\ \Rightarrow m_{CuSO_4}=0,15\times 160=24\ g.\\ b,\\ m_{\text{dd spư}}=m_{CuO}+m_{\text{dd H$_2$SO$_4$}}=12+78,4=90,4\ g.\\ n_{H_2SO_4\text{(dư)}}=0,2-0,15=0,05\ mol.\\ \Rightarrow C\%_{CuSO_4}=\dfrac{24}{90,4}\times 100\%=26,55\%\\ C\%_{H_2SO_4\text{(dư)}}=\dfrac{0,05\times 98}{90,4}\times 100\%=5,42\%\end{array}\)
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