a) Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+56b=22,8\) (1)
Ta có: \(n_{SO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)
Bảo toàn electron: \(2a+3b=0,7\cdot2=1,4\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,25\\b=0,3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25\cdot24}{22,8}\cdot100\%\approx26,32\%\\\%m_{Fe}=73,68\%\end{matrix}\right.\)
b) PTHH: \(2NaOH+H_2SO_{4\left(dư\right)}\rightarrow Na_2SO_4+2H_2O\)
Theo PTHH: \(n_{H_2SO_4\left(dư\right)}=\dfrac{1}{2}n_{NaOH}=\dfrac{0,4\cdot3}{2}=0,6\left(mol\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,15\left(mol\right)\\n_{MgSO_4}=n_{Mg}=0,25\left(mol\right)\end{matrix}\right.\)
Bảo toàn nguyên tố: \(n_{H_2SO_4\left(p/ứ\right)}=n_{MgSO_4}+3n_{Fe_2\left(SO_4\right)_3}+n_{SO_2}=1,4\left(mol\right)\)
\(\Rightarrow\Sigma n_{H_2SO_4}=1,4+0,6=2\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{2\cdot98}{98\%}=200\left(g\right)\)
