a) Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\Rightarrow65a+24b=8,4\left(1\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{HCl}=0,3.2=0,6\left(mol\right)\\n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\end{matrix}\right.\)
PTHH:
`Zn + 2HCl -> ZnCl_2 + H_2`
`Mg + 2HCl -> MgCl_2 + H_2`
Theo PT: `n_{HCl} = 2n_{H_2} = 0,4 (mol) < 0,6 (mol)`
`=> HCl` dư, hh kim loại tan hết
Theo PT: `n_{H_2} = n_{Mg} + n_{Zn}`
`=> a + b = 0,2(2)`
`(1), (2) =>` \(\left\{{}\begin{matrix}a=\dfrac{18}{205}\\b=\dfrac{23}{205}\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{\dfrac{18}{205}.65}{8,4}.100\%=67,94\%\\\%m_{Mg}=100\%-67,94\%=32,06\%\end{matrix}\right.\)
