Question

Answer 1

`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2`

`0,2`     `0,4`             `0,2`       `0,2`        `(mol)`

`n_[Zn]=13/65=0,2(mol)`

`b)V_[H_2]=0,2.22,4=4,48(l)`

`c)m_[dd HCl]=[0,4.36,5]/5 . 100 =292(g)`

`=>C%_[ZnCl_2]=[0,2.136]/[13+292-0,2.2].100~~8,93%`

Answer 2

a: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b: \(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

=>\(n_{H_2}=0.2\left(mol\right)\)

=>\(V_{H_2}=4.48\left(lít\right)\)

Answer 3

a) 

b)