a) 2Al +6HCl \(\rightarrow\)2AlCl3 + 3H2
b)nAl=\(\dfrac{8,1}{27}\)=0,3mol
Theo PT: nH2=\(\dfrac{3}{2}\)nAl=0,45 mol
=> VH2=0,45.22,4=10,08l
c)
Theo PT: nAlCl3=nAl=0,3mol
=> mAlCl3=0,3.133,5=40,05g
Ta có: \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
a) PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2
Theo PTHH và đb, ta có:
a)
\(n_{H_2}=\dfrac{3.0,3}{2}=0,45\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,45.22,4=10,08\left(l\right)\)
b) \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\\ =>m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)