Theo đề bài ta có : nH2 = 0,1(mol)
PTHH :
Zn + 2HCl - > ZnCl2 + H2
0,1mol..0,2mol.....................0,1mol
ZnO + 2HCl - > ZnCl2 + H2O
0,1mol.....0,2mol
a) mZn = 0,1.65 = 6,5(g)
mZnO = 14,6-6,5=8,1(g) => nZnO = 0,1(mol)
b) Ta có :
C%ddHCl(pư) = \(\dfrac{0,4.36,5}{400}.100=3,65\%\)
c) C%ddzncl2 = \(\dfrac{0,2.136}{14,6+400-0,1.2}.100~6,56\%\)
\(\text{a) }n_{H_2}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\left(1\right)\\ ZnO+2HCl\rightarrow ZnCl_2+H_2O\left(2\right)\)
Theo \(pthh\left(1\right):n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Zn}=n\cdot M=0,1\cdot65=6,5\left(g\right)\\ \Rightarrow m_{ZnO}=m_{h^2}-m_{Zn}=14,6-6,5=8,1\left(g\right)\)
b) Theo \(pthh\left(1\right):n_{HCl\left(1\right)}=2n_{H_2}=2\cdot0,1=0,2\left(mol\right)\)
\(n_{ZnO}=\dfrac{m}{M}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
Theo \(pthh\left(2\right):n_{HCl\left(2\right)}=2n_{ZnO}=2\cdot0,1=0,2\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,2+0,2=0,4\left(mol\right)\\ \Rightarrow\Sigma m_{HCl}=n\cdot M=0,4\cdot36,5=14,6\left(g\right)\\ \Rightarrow C_{\%\left(HCl\right)}=\dfrac{m_{ct}}{m_{d^2}}\cdot100=\dfrac{14,6}{400}\cdot100=3,65\%\)
c) Theo \(pthh\left(1\right):n_{ZnCl_2\left(1\right)}=n_{H_2}=0,1\left(mol\right)\)
Theo \(pthh\left(2\right):n_{ZnCl_2\left(2\right)}=n_{ZnO}=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{ZnCl_2}=0,1+0,1=0,2\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=n\cdot M=0,2\cdot136=27,2\left(g\right)\)
Theo \(pthh\left(2\right):n_{H_2O}=n_{ZnO}=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2O}=n\cdot M=0,1\cdot18=1,8\left(g\right)\\ \Rightarrow m_{h^2\left(Sau\text{ }pứ\right)}=1,8+27,2=29\left(g\right)\\ \Rightarrow C_{\%\left(ZnCl_2\right)}=\dfrac{m_{ct}}{m_{d^2}}=\dfrac{27,2}{29}\cdot100=93,79\%\)
