Đáp án:
\(\begin{array}{l} a,\ m_{H_2SO_4}\text{(pư)}=4,9\ g.\\ b,\ m_{CuSO_4}=8\ g.\\ c,\ C\%_{H_2SO_4}\text{(dư)}=9,25\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} PTHH:CuO+H_2SO_4\to CuSO_4+H_2O\\ n_{CuO}=\dfrac{4}{80}=0,05\ mol.\\ n_{H_2SO_4}=\dfrac{49\times 20\%}{98}=0,1\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,05}{1}<\dfrac{0,1}{1}\\ ⇒H_2SO_4\ \text{dư.}\\ Theo\ pt:\ n_{H_2SO_4}\text{(pư)}=n_{CuO}=0,05\ mol.\\ ⇒m_{H_2SO_4}\text{(pư)}=0,05\times 98=4,9\ g.\\ b,\ Theo\ pt:\ n_{CuSO_4}=n_{CuO}=0,05\ mol.\\ ⇒m_{CuSO_4}=0,05\times 160=8\ g.\\ c,\ m_{H_2SO_4}\text{(dư)}=(0,1-0,05)\times 98=4,9\ g.\\ m_{\text{dd spư}}=m_{CuO}+m_{\text{dd H$_2$SO$_4$}}=4+49=53\ g.\\ ⇒C\%_{H_2SO_4}\text{(dư)}=\dfrac{4,9}{53}\times 100\%=9,25\%\end{array}\)
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