a) PTHH : Zn + 2HCl ---> ZnCl2 + H2 (1)
ZnO + 2HCl ---> ZnCl2 + H2O (2)
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{11.2}{22.4}=0,5\left(mol\right)\)
=> nZn = 0.5 mol
=> \(m_{Zn}=n.M=0,5.65=32,5\left(g\right)\)
=> mZnO = 20,25 (g)
b) nZnO = \(\dfrac{m}{M}=\dfrac{20,25}{81}=0,25\left(mol\right)\)
Ta có : \(m_{HCl}=m_{HCl\left(1\right)}+m_{HCl\left(2\right)}\)
= \(M\left(n_{HCl\left(1\right)}+n_{HCl\left(2\right)}\right)=36,5.\left(0,5+1\right)=54,75\left(g\right)\)
=> \(m_{ddddHCl}=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{54,75.100\%}{20\%}=273,75\left(g\right)\)
=> \(m_{H_2O}=273,75-54,75+0,25.18=223,5\left(g\right)\)
Lại có \(m_{ZnCl_2}=M_{ZnCl_2}\left(n_{ZnCl_2\left(1\right)}+n_{ZnCl_2\left(2\right)}\right)\)
= 136(0,5 + 0,25) = 102(g)
\(m_{H_2}=n.M=0,5.2=1\left(g\right)\)
=> \(m_{dd\text{ mới }}=273,75+52,75-1=325,5\)(g)
\(C\%_{ZnCl_2}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{102}{325,5}.100\%=31,34\%\)
\(C\%_{H_2O}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{223,5}{325,5}.100\%=68,66\%\)