\(2Fe\left(a\right)+6H_2SO_4\left(3a\right)\rightarrow Fe_2\left(SO_4\right)_3\left(0,5a\right)+3SO_2\left(1,5a\right)+6H_2O\left(3a\right)\)
\(Cu\left(b\right)+2H_2SO_4\left(2b\right)\rightarrow CuSO_4\left(b\right)+SO_2\left(b\right)+2H_2O\left(2b\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}:a\left(mol\right)\\n_{Cu}:b\left(mol\right)\end{matrix}\right.\)
\(m_{hh}=14,8g\Rightarrow56a+64b=14,8\left(I\right)\)
\(n_{SO_2}=\dfrac{7,28}{22,4}=0,325\left(mol\right)\Rightarrow1,5a+b=0,325\left(II\right)\)
Từ (I);(II) \(\Rightarrow a=0,15;b=0,1\)
\(\Rightarrow\%m_{Fe}\approx56,75\%;\%m_{Cu}=43,25\%\)
b) \(m_{H_2SO_4}=98\left(3.0,15+0,2\right)=63,7\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_460\%}\approx106,167\left(g\right)\)
c) \(m_{ddsaupư}=14,8+106,167-64\left(1,5.0,15+0,1\right)=100,167\left(g\right)\)
\(C\%ddFe_2\left(SO_4\right)_3=\dfrac{30}{100,167}.100\%\approx29,94\%\)
\(C\%ddCuSO_4=\dfrac{16}{100,167}.100\%\approx15,97\%\).
