a)
Ta có : \(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\)
\(2Al + 6HCl \to 2AlCl_3 + 3H_2\)
Theo PTHH :
\(n_{AlCl_3} = n_{Al} = 0,1(mol)\\ \Rightarrow m_{AlCl_3} = 0,1.133,5 = 13,35(gam)\)
b)
\(n_{H_2} = 1,5n_{Al} = 0,15(mol)\\ \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)\)
PTHH: \(2Al+6HCl\)→\(2AlCl_3+6H_2\)
+\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
+\(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)
+\(m_{AlCl_3}=0,1.133,5=13,35\left(gam\right)\)
+\(n_{H_2}=3n_{Al}=0,3\left(mol\right)\)
+\(V_{H_2}=0,3.22,4=6,72\left(mol\right)\)
nAl = \(\dfrac{m}{M}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2 ↑
PT : 2 6 2 3 (mol)
ĐB: 0,1 0,3 0,1 0,15 (mol)
a) mAlCl3 = n.M = 0,1.133,5 = 13,35(g)
b) VH2 = n.22,4 = 0,15.22,4 = 3,36(l)
