\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(2M+6HCl\rightarrow2MCl_3+3H_2\)
\(\dfrac{2}{15}....................0.2\)
\(M_M=\dfrac{11.2}{\dfrac{2}{15}}=84\left(\dfrac{g}{mol}\right)\)
=> Đề sai
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(PTHH:2M+6HCl--->2MCl_3+3H_2\uparrow\)
Theo PT: \(n_M=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,2=\dfrac{2}{15}\left(mol\right)\)
\(\Rightarrow M_M=\dfrac{11,2}{\dfrac{2}{15}}=84\left(\dfrac{g}{mol}\right)\)
Vậy M là Kripton (Kr)
Nếu M hóa trị (II):
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(PTHH:M+2HCl--->MCl_2+H_2\uparrow\)
Theo PT: \(n_M=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow M_M=\dfrac{11,2}{0,2}=56\left(\dfrac{g}{mol}\right)\)
Vậy M là nguyên tố sắt (Fe)
