Question

Answer 1

Câu 2:

PTHH: 3NH₃ + FeCl₃ + 3H₂O ---> 3NH₄Cl + Fe(OH)₃

Ta có: \(n_{NH_3}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(n_{FeCl_3}=1,2.100:1000=0,12\left(mol\right)\)

Ta thấy: \(\dfrac{0,3}{3}< \dfrac{0,12}{1}\)

Vậy FeCl₃ dư.

Theo PT: \(n_{Fe\left(OH\right)_3}=\dfrac{1}{3}.n_{NH_3}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)

=> \(m_{Fe\left(OH\right)_3}=0,1.107=10,7\left(g\right)\)