a) Ta có: \(\left\{{}\begin{matrix}n_{Cu}=\dfrac{0,32}{64}=0,005\left(mol\right)\\n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+27b=0,87-0,32=0,55\) (1)
Bảo toàn electron: \(2a+3b=2n_{H_2}=0,04\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,005\\b=0,01\end{matrix}\right.\)
Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,005\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,005\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow C_{M_{FeSO_4}}=\dfrac{0,005}{0,3}\approx0,02\left(M\right)=C_{M_{Al_2\left(SO_4\right)_3}}\)
b)
Ta thấy trong 0,87 gam hh X có 0,005 mol Fe, 0,005 mol Cu và 0,01 mol Al
\(\Rightarrow\) Trong 2,61 gam hh X có 0,015 mol Fe, 0,015 mol Cu và 0,03 mol Al
PTHH: \(2Fe+6H_2SO_{4\left(đặc\right)}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
\(Cu+2H_2SO_{4\left(đặc\right)}\xrightarrow[]{t^o}CuSO_4+SO_2\uparrow+2H_2O\)
\(2Al+6H_2SO_{4\left(đặc\right)}\xrightarrow[]{t^o}Al_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
Ta có: \(n_{H_2SO_4}=3n_{Fe}+3n_{Al}+2n_{Cu}=0,165\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,165\cdot98}{78\%}\approx20,73\left(g\right)\)