\(n_{H_2SO_4}=0,125.2=0,25\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O
Mol: x 3x x
PTHH: ZnO + H2SO4 → ZnSO4 + H2O
Mol: y y y
C1.
Theo PTHH ta có: \(n_{H_2O}=n_{H_2SO_4}=0,25\Rightarrow m_{H_2O}=0,25.18=4,5\left(g\right)\)
\(m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)
\(\Rightarrow m_{muối}=m_{hh}+m_{H_2SO_4}-m_{H_2O}=16,1+24,5-4,5=36,1\left(g\right)\)
C2.
Ta có: \(\left\{{}\begin{matrix}160x+81y=16,1\\3x+y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
\(\Rightarrow m_{muối}=m_{Fe_2\left(SO_4\right)_3}+m_{ZnSO_4}=0,05.400+0,1.161=36,1\left(g\right)\)
