Question

Answer 1

a)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)

=> 56a + 27b = 22 (1)

\(n_{H_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

             a-->2a-------------->a

            2Al + 6HCl --> 2AlCl₃ + 3H₂

             b-->3b---------------->1,5b

=> a + 1,5b = 0,8 (2)

(1)(2) => a = 0,2 (mol); b = 0,4 (mol)

\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2.56}{22}.100\%=50,91\%\\\%m_{Al}=\dfrac{0,4.27}{22}.100\%=49,09\%\end{matrix}\right.\)

b)

m_HCl = (2a + 3b).36,5 = 58,4 (g)

=> \(m_{dd.HCl}=\dfrac{58,4.100}{7,3}=800\left(g\right)\)