a) 2Al + 6HCl --> 2AlCl3 + 3H2
b) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,1->0,3---->0,1---->0,15
=> VH2 = 0,15.22,4 = 3,36 (l)
c) mAlCl3 = 0,1.133,5 = 13,35 (g)
d) \(V_{dd.HCl}=\dfrac{0,3}{2}=0,15\left(l\right)\)
a) Ta có PT : 2Al + 6HCl ----> 2AlCl3 + 3H2
b) nAl = 2,7\27=0,1(mol)
Theo PT ta có: nHCl = 3nAl = 0,1 . 3 = 0,3(mol)
=>VHCl=\(\dfrac{0,3}{2}\)=0,15 l=150ml
c) Theo PT ta có: nAlCl3=nAl = 0,1(mol)
=> mAlCl3 = 0,1 . 133,5 = 13.35(g)
c) Theo PT ta có: nH2= =0,15(mol)
=> VH2= 0,15 . 22,4 = 3,36 (l)