giúp với ạ
PTHH: \(2R+2xHCl\rightarrow2RCl_x+xH_2\uparrow\) (x là hóa trị của R)
a) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow n_R=\dfrac{0,2}{x}\left(mol\right)\) \(\Rightarrow M_R=\dfrac{6,5}{\dfrac{0,2}{x}}=\dfrac{65x}{2}\)
Ta thấy với \(x=2\) thì \(M_R=65\) (Kẽm)
b+c) Theo PTHH: \(\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{ZnCl_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2\cdot36,5}{7,3\%}=100\left(g\right)\\m_{ZnCl_2}=136\cdot0,1=13,6\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Zn}+m_{ddHCl}-m_{H_2}=106,3\left(g\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{13,6}{106,3}\cdot100\%\approx12,8\%\)
