Question

Answer 1

4,2 

a,PTHH: Fe + 2HCl------> FeCl2 + H2 

b, nFe=0,4 mol

\(\Rightarrow\) nH2=0,4 mol 

\(\Rightarrow\) vH2=8,96 l

c, theo pt 

nFeCl2/nFe=1/1 

\(\Rightarrow\) nFeCl2=0,4 mol 

\(\Rightarrow\) mFeCl2=50,8 g

 

Answer 2

Câu 4:

4.1/ Ta có: \(n_{NaCl}=2,5.0,9=2,25\left(mol\right)\)

\(\Rightarrow m_{NaCl}=2,25.58,5=131,625\left(g\right)\)

4.2/ Ta có: \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

_____0,4___________0,4___0,4 (mol)

b, \(V_{H_2}=0,4.22,4=8,96\left(l\right)\)

c, \(m_{FeCl_2}=0,4.127=50,8\left(g\right)\)

Bạn tham khảo nhé!