Cho hỗn hợp Al và Fe có tỉ lệ số mol là 3:5 tác dụng hết với dung dịch HCl dư, sau phản ứng thu được 21,28 lít khí ở đktc. Tính khối lượng mỗi kim loại trong hỗn hợp ban đầu.

2 câu trả lời

Bạn tham khảo:

$2Al+6HCl \to 2AlCl_3+3H_2$
$Fe+2HCl \to FeCl_2+H_2$
$n_{Al}=3a$
$n_{Fe}=5a$

$n_{H_2}=\dfrac{21,28}{22,4}=0,95(mol)$
$n_{H_2}=4,5a+5b=0,95$
$\to a=0,1$

$m_{Al}=0,1.27.3=8,1(g)$

$m_{Fe}=0,1.56.5=28(g)$

Đáp án:

\(m_{Al}=8,1g\\m_{Fe}=28g\) 

Giải thích các bước giải:

\(PTHH :\\2Al + 6HCl \to 2AlCl_3 + 3H_2\\3a \qquad \qquad \qquad \quad \to \quad 4,5a \qquad (mol)\\Fe + 2HCl \to FeCl_2 + H_2\\5a \qquad \qquad \quad \quad \quad \to \quad 5a \qquad \quad (mol)\\\text{Gọi} \ n_{Al}=3a \ \ (mol) \to n_{Fe}=5a \ \ (mol)\\\text{Ta có :} \ n_{H_2}=\dfrac{21,28}{22,4}=0,95 \ \ (mol) \\⇔ 4,5a+5a=0,95\\⇔ a=0,1\\\to \begin{cases}m_{Al}=0,1.3.27=8,1 \ \ (g)\\ m_{Fe}=0,1.5.56=28 \ \ (g)\end{cases}\)

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