a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\left(1\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(\left(1\right)\Rightarrow n_{HCl}=2.0,2=0,4\left(mol\right)\)
\(m_{dd}\left(HCl\right)=\dfrac{0,4.36,5}{20}.100=73\left(g\right)\)
b) \(\left(1\right)\Rightarrow n_{ZnCl_2}=n_{H_2}=0,2\left(mol\right)\)
\(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
\(m_{H_2}=0,2.2=0,4\left(g\right)\)
\(m_{dd}\left(sau.pư\right)=m_{Zn}+m_{dd}\left(HCl\right)-m_{H_2}=13+73-0,4=85,6\left(g\right)\)
\(C\%\left(ZnCl_2\right)=\dfrac{27,2}{85,6}.100\%=31,78\%\)
Đúng 2
Bình luận (0)