Question

Giúp với tối cần gấp

Answer 1

Fe + 2HCl \(\rightarrow\)FeCl₂ +H₂

n_Fe=\(\dfrac{2,8}{56}=0,05\left(mol\right)\)

Theo PTHH ta có:

2n_Fe=n_HCl=0,1(mol)

V_{dd HCl}=\(\dfrac{0,1}{2}=0,05\left(lít\right)\)

b;

Theo PTHH ta có:

n_Fe=n_H2=n_FeCl2=0,05(mol)

V_H2=0,05.22,4=1,12(lít)

c;

C_M dd FeCl₂=\(\dfrac{0,05}{0,05}=1M\)