n\(_{Al}\) = \(\dfrac{m}{M}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
2Al + 3H\(_2\)SO\(_4\) → \(Al_2\left(SO_4\right)_3\) + 2H\(_2\)
(mol) 0,1 → 0,1
Thể tích khí Hiđrô thoát ra:
\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)
b)
n\(_{HCl}=\dfrac{8,1}{36,5}=\dfrac{81}{365}\left(mol\right)\)
2Al + 6HCl → 2AlCl\(_3\) + 3H\(_2\)
2 mol 6 mol
0,1 mol \(\dfrac{81}{365}\left(mol\right)\)
Tỉ lệ: \(\dfrac{0,1}{2}>\dfrac{\dfrac{81}{365}}{6}\Rightarrow Al\)
2Al + 6HCl → 2AlCl\(_3\) + 3H\(_2\)
(mol) \(\dfrac{27}{365}\) ← \(\dfrac{81}{365}\)
m\(_{Al\left(pư\right)}\) = n . M = \(\dfrac{27}{365}\) . 27 = \(\dfrac{729}{365}\left(g\right)\)
m\(_{Al\left(dư\right)}=m_{Al\left(bđ\right)}-m_{Al\left(pư\right)}\)
= 2,7 - \(\dfrac{729}{365}\)
= \(\dfrac{513}{730}\left(g\right)\)
Vậy Al dư và dư \(\dfrac{513}{730}g\)
