Question

Answer 1

\(n_{Zn}=\dfrac{13}{65}=0,2mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,2                                    0,2  ( mol )

\(V_{H_2}=0,2.22,4=4,48l\)

--> Chọn C

Answer 2

4,48

Answer 3

4,48 ó bẹn

Answer 4

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           0,2---------------------->0,2

=> x = 0,2.22,4 = 4,48 (l)

=> C

Answer 5

= 4,48

Answer 6

4,48