Question

Giúp e câu 382

Answer 1

Gọi số mol Al, Cu, Fe là a, b , c

=> 27a + 64b + 56c = 17,5

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

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

Fe + 2HCl --> FeCl₂ + H₂

c------------------------>c

=>1,5a + c = 0,3

PTHH: 2Al + 3Cl₂ --t^o--> 2AlCl₃
_____a------------------>a

2Fe + 3Cl₂ --t^o--> 2FeCl₃

c------------------->c

Cu + Cl₂ --t^o--> CuCl₂

b---------------->b

=> 133,5a + 135b + 162,5c = 51,225

=> a = 0,1; b = 0,1; c = 0,15

=> \(\%Fe=\dfrac{56.0,15}{17,5}.100\%=48\%\)

=> A