a) PTHH : \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
b) \(n_{Al}=\dfrac{m}{M}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(\Rightarrow n_{H_2}=\dfrac{0,1\times3}{2}=0,15\left(mol\right)\)
\(\Rightarrow V_{H_{2\left(ĐKTC\right)}}=n.22,4=3,36\left(l\right)\)
\(n_{HCl}=\dfrac{0,1\times6}{2}=0,3\left(mol\right)\)
\(\Rightarrow m_{HCl}=n.M=10,95\left(g\right)\)
\(n_{AlCl\left(3\right)}=n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=n.M=13,35\left(g\right)\)
a) 2Al + 6HCl -> 2AlCl3 + 3H2
b)nAl = 2.7/27=0.1mol
2Al + 6HCl -> 2AlCl3 + 3H2
(mol) 0.1 0.3 0.1 0.15
VH2 = 0.15*22.4=3.36l
mHCl = 0.3*36.5=10.95g
mAlCl3= 0.1*133.5=13.35g
