Question

Giúp mình vớ

Answer 1

a) Gọi số mol Al, Fe là a, b (mol)

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

\(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

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

            Fe + H₂SO₄ --> FeSO₄ + H₂

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

=> 1,5a + b = 0,04 (2)

(1)(2) => a = 0,02 (mol); b = 0,01 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,02.27}{1,1}.100\%=49,09\%\\\%m_{Fe}=\dfrac{0,01.56}{1,1}.100\%=50,91\%\end{matrix}\right.\)

b) n_H2SO4 = 1,5a + b = 0,04 (mol)

=> \(C\%_{dd.H_2SO_4}=\dfrac{0,04.98}{49}.100\%=8\%\)