2Al + 6HCl -> 2AlCl3 + 3H2
Ta có:
nH2 = 13,4 / 22,4 = 0,6 mol
=> nAl = 2/3 . nH2 = 2/3 . 0,6= 0,4 mol
-> mAl = 0,4 . 27 = 10,8 g
+) nHCl = 2 . nH2 = 2. 0,6 = 1,2 mol
=> mHCl = 1,2 . 36,5 = 43,8 g
+) nAlCl3 = 2/3 . nH2 = 2/3 . 0,6 = 0,4 mol
=> mAlCl3 = 0,4 . 133,5 = 53,4 g
Vậy...
2Al + 6HCl → 2AlCl3 + 3H2
\(n_{H_2}=\dfrac{13,4}{22,4}=\dfrac{67}{112}\left(mol\right)\)
a) Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}\times\dfrac{67}{112}=\dfrac{67}{168}\left(mol\right)\)
\(\Rightarrow m_{Al}=\dfrac{67}{168}\times27=10,77\left(g\right)\)
b) Theo PT: \(n_{HCl}=2n_{H_2}=2\times\dfrac{67}{112}=\dfrac{67}{56}\left(mol\right)\)
\(\Rightarrow m_{HCl}=\dfrac{67}{56}\times36,5=43,67\left(g\right)\)
c) Theo PT: \(n_{AlCl_3}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}\times\dfrac{67}{112}=\dfrac{67}{168}\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=\dfrac{67}{168}\times133,5=53,24\left(g\right)\)
