Question

Answer 1

a)

2Al + 6HCl --> 2AlCl₃ + 3H₂

Fe + 2HCl --> FeCl₂ + H₂

b) Gọi số mol Al, Fe là a, b

=> 27a + 56b = 2,22

\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

______a---->3a--------------->1,5a

Fe + 2HCl --> FeCl₂ + H₂

b---->2b--------------->b

=> 1,5a + b = 0,06

=> a = 0,02; b = 0,03

=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,02.27}{2,22}.100\%=24,324\%\\\%Fe=\dfrac{0,03.56}{2,22}.100\%=75,676\%\end{matrix}\right.\)

c) n_HCl = 3a + 2b = 0,12(mol)

=> \(C_{M\left(ddHCl\right)}=\dfrac{0,12}{0,2}=0,6M\)