Question

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Answer 1

a, Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

b, \(n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ n_{H_2}=0,2mol\\ V_{H_2}=0,2.22,4=4,48l\)

\(c,n_{FeCl_2}=0,2mol\\ m_{FeCl_2}=0,2.\left(56+35,5.2\right)=25,4g\)

Answer 2

a. PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b. \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

1 mol Fe tạo ra 1 mol H₂ 

=> 0,2 mol Fe tạo ra 0,2 mol H₂

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c. 

1 mol Fe tạo ra 1 mol FeCl₂.

=> 0,2 mol Fe tạo ra 0,2 mol FeCl₂

\(m_{FeCl_2}=0,2.\left(56+35,5.2\right)=25,4\left(g\right)\)