2Al + 6HCl → 2AlCl3 + 3H2↑ (1)
\(n_{Al}=\frac{2,7}{27}=0,1\left(mol\right)\)
\(n_{HCl}=0,1\times6=0,6\left(mol\right)\)
Theo PT1: \(n_{Al}=\frac{1}{3}n_{HCl}\)
Theo bài: \(n_{Al}=\frac{1}{6}n_{HCl}\)
Vì \(\frac{1}{6}< \frac{1}{3}\) ⇒ HCl dư
a) Theo PT1: \(n_{H_2}=\frac{3}{2}n_{Al}=\frac{3}{2}\times0,1=0,15\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,15\times22,4=3,36\left(l\right)\)
b) HCl + NaOH → NaCl + H2O (2)
Theo pT1: \(n_{HCl}pư=3n_{Al}=3\times0,1=0,3\left(mol\right)\)
\(\Rightarrow n_{HCl}dư=0,6-0,3=0,3\left(mol\right)\)
Theo PT2: \(n_{NaOH}=n_{HCl}dư=0,3\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,3\times40=12\left(g\right)\)
nAl= 2.7/27= 0.1 mol
nHCl = 0.1*6 = 0.6 mol
2Al + 6HCl --> 2AlCl3 + 3H2
Bđ: 0.1___0.6
Pư : 0.1___0.3_____________0.15
Kt: 0_____0.3______________0.15
VH2= 0.15*22.4=3.36l
NaOH + HCl --> NaCl + H2O
0.3_____0.3
mNaOH = 0.3*40=12g
