\(n_{H_2}=\frac{11,2}{22,4}=0,5\left(Mol\right)\)
Gọi số mol Fe, Al lần lượt là a,b
PTHH: Fe + H2So4 --> FeSO4 + H2
______ a ----> a ---------> a-------> a (mol)
2Al + 3H2SO4 --> Al2(SO4)3 + 3h2
b --------> 1,5b -------------------> 1,5b (mol)
=> a + 1,5b = 0,5 (1)
- \(m_{H2SO4}=\left(a+1,5b\right).98=98a+147b\) (g)
=> m dd H2SO4 = \(\frac{\left(98a+147b\right)100}{19,6}=500a+750b\left(g\right)\)
\(m_{FeSO_4}=152a\left(g\right)\)
Ta có: C% FeSO4 = \(\frac{152a}{500a+750b+56a+27b}.100\%=11,44578313\%\)
=> a=b (2)
(1)(2) => \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)