`n_R=\frac{16,2}{R}(mol)`
Bảo toàn khối lượng:
`m_{Cl_2}=m_{D}-m_{R}`
`=> m_{Cl_2}=58,8-16,2=42,6g`
`=> n_{Cl_2}=\frac{42,6}{71}=0,6(mol)`
`2R+nCl_2\to 2RCl_n`
`=> n_{R}=\frac{1,2}{n}(mol)(1)`
Bảo toàn khối lượng:
`m_{O_2}=m_{E}-m_{D}`
`=> m_{O_2}=63,6-58,8=4,8g`
`=> n_{O_2}=\frac{4,8}{32}=0,15(mol)`
`4R+nO_2\overset{t^o}{\to}2R_2O_n`
`=> n_R=\frac{0,6}{n}(mol)(2)`
Ta có: `n_R=(1)+(2)`
`=> \frac{16,2}{R}=\frac{0,6}{n}+\frac{1,2}{n}`
`=>\frac{16,2}{R}=\frac{1,8}{n}`
`=> 1,8R=16,2n`
`=>R=9n`
`=> n=3 \text{(t/m)}`
Vậy `R` là `Al`.
`=> n_{AlCl_3}=0,4(mol) => %m_{AlCl_3}=\frac{133,5.0,4.100%}{63,6}\approx 84%`
`=> %m_{Al_2O_3}=100%-84%=16%`