\(n_{H_2SO_4}=0.1\cdot0.25=0.025\left(mol\right)\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(TừPTHH:\)
\(n_{H_2SO_4}=n_{H_2O}=0.025\left(mol\right)\)
\(BTKL:\)
\(m_{Muối}=1.405+0.025\cdot98-0.025\cdot18=3.405\left(g\right)\)