Đáp án:

Câu 5.1:

\(\begin{array}{l} C_{M_{\text{dd spư}}}=C_{M_{MgSO_4}}=0,5\ M.\end{array}\)

Câu 5.2:

\(\begin{array}{l} a,\ a=m_{Fe}=22,4\ g.\\ b,\ C\%_{\text{dd HCl}}=7,3\%\end{array}\)

Giải thích các bước giải:

Câu 5.1:

\(\begin{array}{l} PTHH:MgO+H_2SO_4\to MgSO_4+H_2O\\ n_{MgO}=\dfrac{8}{40}=0,2\ mol.\\ Theo\ pt:\ n_{MgSO_4}=n_{MgO}=0,2\ mol.\\ V_{\text{dd spư}}=V_{\text{dd tpư}}=400\ ml=0,4\ lít.\\ \Rightarrow C_{M_{\text{dd spư}}}=C_{M_{MgSO_4}}=\dfrac{0,2}{0,4}=0,5\ M.\end{array}\)

Câu 5.2:

\(\begin{array}{l} a,\\ PTHH:\\ Fe+2HCl\to FeCl_2+H_2\uparrow\\ n_{H_2}=\dfrac{0,8}{2}=0,4\ mol.\\ Theo\ pt:\ n_{Fe}=n_{H_2}=0,4\ mol.\\ \Rightarrow a=m_{Fe}=0,4\times 56=22,4\ g.\\ b,\\ \text{Áp dụng ĐLBT khối lượng ta có:}\\ m_{Fe}+m_{\text{dd HCl}}=m_{\text{dd muối}}+m_{H_2}\\ \Rightarrow m_{\text{dd HCl}}=m_{\text{dd muối}}+m_{H_2}-m_{Fe}\\ \Rightarrow m_{\text{dd HCl}}=421,6+0,8-22,4=400\ g.\\ Theo\ pt:\ n_{HCl}=2n_{H_2}=0,8\ mol.\\ \Rightarrow C\%_{\text{dd HCl}}=\dfrac{0,8\times 36,5}{400}\times 100\%=7,3\%\end{array}\)

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