PTHH: \(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\) (1)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\) (2)
a) Ta có: \(n_{H_2SO_4}=0,4\cdot2=0,8\left(mol\right)\)
Đặt số mol của FeO là \(a\) \(\Rightarrow n_{H_2SO_4\left(1\right)}=a\)
Đặt số mol của MgO là \(b\) \(\Rightarrow n_{H_2SO_4\left(2\right)}=b\)
Ta có hệ phương trình:
\(\left\{{}\begin{matrix}72a+40b=48\\a+b=0,8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=0,3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{FeO}=0,5mol\\n_{MgO}=0,3mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeO}=72\cdot0,5=36\left(g\right)\\m_{MgO}=48-36=12\left(g\right)\end{matrix}\right.\)
b) Theo 2 PTHH: \(\left\{{}\begin{matrix}n_{FeO}=n_{FeSO_4}=0,5mol\\n_{MgO}=n_{MgSO_4}=0,3mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,5\cdot152=76\left(g\right)\\m_{MgSO_4}=0,3\cdot120=36\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{ddH_2SO_4}=400\cdot1,2=480\left(g\right)\) \(\left(D=1,2g/ml\right)\)
\(\Rightarrow m_{dd}=m_{oxit}+m_{ddH_2SO_4}=48+480=528\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\frac{76}{528}\cdot100\approx14,39\%\\C\%_{MgSO_4}=\frac{36}{528}\cdot100\approx6,82\%\end{matrix}\right.\)