Question

giúp em c3 với ạ

Answer 1

Câu 3:

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)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,02.27}{1,1}.100\%=49,1\%\\\%m_{Fe}=\dfrac{0,01.56}{1,1}.100\%=50,9\%\end{matrix}\right.\)

b) \(m_{H_2SO_4\left(lý.thuyết\right)}=\left(1,5a+b\right).98=3,92\left(g\right)\)

=> \(m_{H_2SO_4\left(tt\right)}=\dfrac{3,92.120}{100}=4,704\left(g\right)\)

=> \(C\%=\dfrac{4,704}{49}.100\%=9,6\%\)

c) \(\left\{{}\begin{matrix}n_{Al}=\dfrac{0,66.49,1\%}{27}=0,012\left(mol\right)\\n_{Fe}=\dfrac{0,66.50,9\%}{56}=0,006\left(mol\right)\end{matrix}\right.\)

PTHH: 2Al + 6H₂SO₄ --> Al₂(SO₄)₃ + 3SO₂ + 6H₂O

          0,012------------------------->0,018

            2Fe + 6H₂SO₄ --> Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

           0,006------------------------->0,009

=> V_SO2 = (0,018 + 0,009).22,4 = 0,6048 (l)