2M + 2nHCl → 2MCln + nH2
\(n_{H_2}=\frac{1,344}{22,4}=0,06\left(mol\right)\)
\(n_M=\frac{1,44}{M_M}\left(mol\right)\)
a) Theo PT: \(n_M=\frac{2}{n}n_{H_2}=\frac{0,12}{n}\left(mol\right)\)
\(\Leftrightarrow\frac{1,44}{M_M}=\frac{0,12}{n}\)
\(\Rightarrow M_M=\frac{1,44n}{0,12}\)
Lập bảng:
| n | 1 | 2 | 3 |
| MM | 12 | 24 | 36 |
| loại | Mg | loại |
Vậy M là Mg
b) PTHH: Mg + 2HCl → MgCl2 + H2
Theo PT: \(n_{HCl}=2n_{H_2}=0,12\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,12\times36,5=4,38\left(g\right)\)
\(\Rightarrow C\%_{ddHCl}=\frac{4,38}{58,4}\times100\%=7,5\%\)
c) \(m_{H_2}=0,06\times2=0,12\left(g\right)\)
\(m_{dd}saupư=1,44+58,4-0,12=59,72\left(g\right)\)
Theo PT: \(n_{MgCl_2}=n_{H_2}=0,06\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,06\times95=5,7\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\frac{5,7}{59,72}\times100\%=9,54\%\)
