Đáp án:
\(\begin{array}{l} b,\ C\%_{MgSO_4}=14,89\%\\ C\%_{Fe_2(SO_4)_3}=4,96\%\\ C\%_{H_2SO_4\text{(dư)}}=7,3\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\\ PTHH:\\ MgO+H_2SO_4\to MgSO_4+H_2O\ (1)\\ Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\ (2)\\ b,\\ m_{MgO}=11,2\times 71,43\%=8\ g.\\ m_{Fe_2O_3}=11,2-8=3,2\ g.\\ m_{H_2SO_4}=150\times 25\%=37,5\ g.\\ n_{MgO}=\dfrac{8}{40}=0,2\ mol.\\ n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02\ mol.\\ n_{H_2SO_4}=\dfrac{37,5}{98}≈0,38\ mol.\\ \sum n_{H_2SO_4\text{(phản ứng)}}=n_{MgO}+3n_{Fe_2O_3}=0,2+0,02\times 3=0,26\ mol.\\ \Rightarrow n_{H_2SO_4\text{(dư)}}=0,38-0,26=0,12\ mol.\\ m_{\text{dd spư}}=m_{\text{hỗn hợp}}+m_{\text{dd H$_2$SO$_4$}}\\ \Rightarrow m_{\text{dd spư}}=11,2+150=161,2\ g.\\ Theo\ pt\ (1):\ n_{MgSO_4}=n_{MgO}=0,2\ mol.\\ Theo\ pt\ (2):\ n_{Fe_2(SO_4)_3}=n_{Fe_2O_3}=0,02\ mol.\\ \Rightarrow C\%_{MgSO_4}=\dfrac{0,2\times 120}{161,2}\times 100\%=14,89\%\\ C\%_{Fe_2(SO_4)_3}=\dfrac{0,02\times 400}{161,2}\times 100\%=4,96\%\\ C\%_{H_2SO_4\text{(dư)}}=\dfrac{0,12\times 98}{161,2}\times 100\%=7,3\%\end{array}\)
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