\(n_A=\dfrac{5,4}{M_A}\left(mol\right)\)
PTHH: 2xA + yO2 --to--> 2AxOy
_____\(\dfrac{5,4}{M_A}\) ------------->\(\dfrac{5,4}{x.M_A}\)
=> \(\dfrac{5,4}{x.M_A}\left(x.M_A+16y\right)=10,2\)
=> \(M_A=9.\dfrac{2y}{x}\)
Xét \(\dfrac{2y}{x}=3=>M_A=27\left(Al\right)=>\dfrac{x}{y}=\dfrac{2}{3}=>Al_2O_3\)
\(2xA+yO_2\overset{t^o}{--->}2A_xO_y\)
Áp dụng ĐLBTKL, ta có:
\(m_A+m_{O_2}=m_{A_xO_y}\)
\(\Leftrightarrow5,4+m_{O_2}=10,2\)
\(\Leftrightarrow m_{O_2}=10,2-5,4=4,8\left(g\right)\)
\(\Rightarrow n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)
Theo PT: \(n_{A_xO_y}=\dfrac{2}{y}.n_{O_2}=\dfrac{2}{y}.0,15=\dfrac{0,3}{y}\left(mol\right)\)
\(\Rightarrow n_{A_{\left(A_xO_y\right)}}=\dfrac{0,3}{y}.x=\dfrac{0,3x}{y}\left(mol\right)\)
Theo PT: \(n_A=\dfrac{2x}{y}.n_{O_2}=\dfrac{0,3x}{y}\left(mol\right)\)
\(\Rightarrow\dfrac{0,3x}{y}.A=5,4\)
\(\Leftrightarrow\dfrac{A}{9}=\dfrac{2y}{x}\)
Biện luận:
| 2y/x | 1 | 2 | 3 |
| A | 9 | 18 | 27 |
| loại | loại | Al |
Vậy A là nhôm (Al)
