Question

Giúp t 2 câu nì

Answer 1

Câu III

1) \(\left\{{}\begin{matrix}\%Fe=\dfrac{56.2}{160}.100\%=70\%\\\%O=\dfrac{16.3}{160}.100\%=30\%\end{matrix}\right.\)

2)

a) Fe + 2HCl --> FeCl₂ + H₂

b) PTHH: Fe + 2HCl --> FeCl₂ + H₂

_______0,5--->1

=> m_HCl = 1.36,5 = 36,5(g)

Câu IV

\(n_A=\dfrac{1}{22,4}=\dfrac{5}{112}\left(mol\right)=>M_A=\dfrac{1,9643}{\dfrac{5}{112}}=44\left(g/mol\right)\)

CTHH: N_xO_y

=> 14x + 16y = 44

Xét x = 1 => y = \(\dfrac{15}{8}\left(L\right)\)

Xét x = 2 => y = 1=> CTHH: N₂O

Answer 2

\(1.\%Fe=\dfrac{112}{160}=70\%\\ \%O=100\%-70\%=30\%\)

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