Bài 1:
PTHH: \(X+2HCl\rightarrow XCl_2+H_2\uparrow\)
Ta có: \(n_X=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\cdot\left(0,2\cdot0,3\right)=0,03\left(mol\right)\)
\(\Rightarrow M_X=\dfrac{1,95}{0,03}=65\) (Kẽm)
Theo PTHH: \(n_{ZnCl_2}=0,03\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,03\cdot136=4,08\left(g\right)\\C_{M_{ZnCl_2}}=\dfrac{0,03}{0,2}=0,15\left(M\right)\end{matrix}\right.\)
Bài 2:
PTHH: \(2A+2xHCl\rightarrow2ACl_x+xH_2\uparrow\) (x là hóa trị của kim loại A)
Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\) \(\Rightarrow n_A=\dfrac{0,1}{x}\left(mol\right)\) \(\Rightarrow M_A=\dfrac{1,2}{\dfrac{0,1}{x}}=12x\)
Ta thấy với \(x=2\) thì \(M_A=24\) (Kim loại Magie)
Theo PTHH: \(n_{MgCl_2}=0,05\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,05\cdot95=4,75\left(g\right)\\C_{M_{MgCl_2}}=\dfrac{0,05}{0,5}=0,1\left(M\right)\end{matrix}\right.\)
