a) Gọi \(\left\{{}\begin{matrix}n_{Fe_2O_3}=a\left(mol\right)\\n_{MgO}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
a--------->3a---------->a
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
b-------->b----------->b
b) Theo bài ra, ta có hệ: \(\left\{{}\begin{matrix}160a+40b=18\\400a+120b=46\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=0,1.160=16\left(g\right)\\m_{MgO}=0,05.40=2\left(g\right)\end{matrix}\right.\)
c) \(m_{H_2SO_4\left(TT\right)}=\left(3.0,1+0,05\right).98=34,3\left(g\right)\)
\(\rightarrow m_{H_2SO_4\left(LT\right)}=34,3+\dfrac{15}{100}.34,3=39,445\left(g\right)\)