`a,m_(ddHCl)=82,2.1,2=98,64(g)`
`=>m_(HCl)=\frac{98,64.37}{100}=36,5(g)`
`=>n_(HCl)=\frac{36,5}{36,5}=1(mol)`
`n_(H_2)=\frac{7,84}{22,4}=0,35(mol)`
`2M+ 2nHCl-> 2MCl_n+ nH_2`
`\frac{0,7}{n}` `0,35`
`=> M_M=\frac{8,4}{\frac{0,7}{n}} =12n`
`\text{Với}` `+n=1 => M=12` `\text{(g/mol)(loại)}`
`+n=2 => M=24` `\text{(g/mol)(Mg)}`
`+n=3 => M=36` `\text{(g/mol)(loại)}`
`\text{Vậy M là Mg}`
`b,`
`\text{PTHH}`
`Mg+ 2HCl-> MgCl_2+ H_2`
`0,7` `0,35` `0,35`
`n_(\text{HCl dư})=1-0,7=0,3(mol)`
`=> m_(HCl)=0,3.36,5=10,95(g)`
`m_(MgCl_2)=0,35.95=33,25(g)`
`m_(H_2)=0,35.2=0,7(g)`
`m_(dd)=8,4+98,64-0,7=106,34(g)`
`=> C%_(HCl)=\frac{10,95}{106,34}.100=10,3%`
`C%_(MgCl_2)=\frac{33,25}{106,34}.100=31,27%`
`\text{Đề là 7,8l}` `H_2` `\text{nhưng mình lấy 7,84 l}` `H_2` `\text{nha vì nếu để 7,8 thì số lẻ quá :3}`
