Đặt số mol Al2O3, Fe3O4 lần lượt là x, y (mol).
\(Al_2O_3+3H_2SO_4\rightarrow2Al_2\left(SO_4\right)_3+3H_2O\)
\(2Fe_3O_4+10H_2SO_4\rightarrow3Fe_2\left(SO_4\right)_3+SO_2+10H_2O\)
\(n_{H2SO4}=\dfrac{m_{\text{dd}}.C\%}{M.100\%}=\dfrac{44.20\%}{98.100\%}=\dfrac{22}{245}\left(mol\right)\)
Theo đề ta có hệ phương trình:
\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\3x+5y=\dfrac{22}{245}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{44}{5145}\left(mol\right)\\y=\dfrac{22}{1715}\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m=m_{Al}+m_{Fe}=\dfrac{44}{5145}.27+\dfrac{22}{1715}.56=\dfrac{1628}{1715}\approx0,95\left(g\right)\)
