a)2Al + 6HCl---->2AlCl3 +3H2
b) Ta có
n\(_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)
n\(_{HCl}=0,744.1-0,744\left(mol\right)\)
=> HCl dư
Theo pthh
n\(_{H2}=\frac{3}{2}n_{Al}=0,3\left(mol\right)\)
V\(_{H2}=0,3.22,4=6.72\left(l\right)\)
c) Theo pthh
n\(_{HCl}=3n_{Al}=0,6\left(mol\right)\)
n\(_{HCl}dư=0,744-0,6=0,144\left(mol\right)\)
C\(_M=\frac{0,144}{0,744}=0,19\left(M\right)\)
Theo pthh
n\(_{AlCl3}=n_{Al}=0,2\left(mol\right)\)
C\(_M=\frac{0,2}{0,744}=0,27\left(M\right)\)
\(\text{nHCl=0,744.1=0,744(mol) }\)
\(nHCl>nAl\)
=>HCl dư ,Al hết
Vì \(Al=0,2\Rightarrow nH2=0,3\)
\(\Rightarrow VH2=0,3.22,4=6,72l\)
\(\Rightarrow CMAl2O3\approx0,27\)
\(\Rightarrow CM_{HCl}dư\approx0,19M\)
nAl = 5,4/27 = 0,2 mol
nHCl = 0,744.1 = 0,744 mol
2Al + 6HCl --> 2AlCl3 +3 H2
0,2 0,6 0,2 0,3
Ta có: \(\frac{0,2}{2}< \frac{0,744}{6}\) => nHCl dư
VH2 = 0,3.22,4 = 6,72 l
nHCl dư = 0,744 - 0,6 = 0,144 mol
CM HCl dư = 0,144/0,744 = 0,19 M
CM AlCl3 = 0,2/0,744 = 0,27 M
