Question

Answer 1

Câu 14 :

a. \(n_{Zn}=\dfrac{16}{65}\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

             \(\dfrac{16}{65}\)      \(\dfrac{32}{65}\)                     \(\dfrac{16}{65}\)

b. \(V_{H_2}=\dfrac{16}{65}.22,4=\dfrac{358.4}{65}\left(l\right)\)

c. \(m_{HCl}=\dfrac{32}{65}.36,5=\dfrac{1168}{65}\left(g\right)\)

Answer 2

bài 14 nha

Answer 3

nZn = 16/65 (mol)

PTHH: Zn + 2HCl -> ZnCl2 + H2

Mol: 16/65 ---> 32/65 ---> 16/65 ---> 16/65

VH2 = 16/65 . 22,4 = 1792/325 (l)

mHCl = 32/65 . 36,5 = 1168/65 (g)