2Al + 6HCl → 2AlCl3 + 3H2 (1)
Al2O3 + 6HCl → 2AlCl3 + 3H2O (2)
\(n_{H_2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
a) Theo pT: \(n_{Al}=\frac{2}{3}n_{H_2}=\frac{2}{3}\times0,6=0,4\left(mol\right)\)
\(m_{Al}=0,4\times27=10,8\left(g\right)\)
\(\Rightarrow m_{Al_2O_3}=21-10,8=10,2\left(g\right)\)
\(\%m_{Al}=\frac{10,8}{21}\times100\%=51,43\%\)
\(\%m_{Al_2O_3}=100\%-51,43\%=48,57\%\)
b) Theo Pt1: \(n_{HCl}=3n_{Al}=3\times0,4=1,2\left(mol\right)\)
\(n_{Al_2O_3}=\frac{10,2}{102}=0,1\left(mol\right)\)
Theo pt2: \(n_{HCl}=6n_{Al_2O_3}=6\times0,1=0,6\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=1,2+0,6=1,8\left(mol\right)\)
\(\Rightarrow\Sigma m_{HCl}=1,8\times36,5=65,7\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{65,7}{36\%}=182,5\left(g\right)\)
\(\Rightarrow V_{ddHCl}=\frac{182,5}{1,18}=154,66\left(ml\right)\)
Đặt :
nAl = x mol
nAl2O3 = y mol
mhh= 27x + 102y = 21 g (1)
nH2= 13.44/22.4=0.6 mol
2Al + 6HCl --> 2AlCl3 + 3H2
0.4____1.2_____________0.6
Từ (1) :
27*0.4 + 102y = 21
=> y = 0.1
Al2O3 + 6HCl --> 2AlCl3 + 3H2O
0.1______0.6
mAl2O3 = 10.2 g
mAl = 10.8 g
%Al=51.42%
%Al2O3=48.58%
nHCl = 1.2 + 0.6 = 1.8 mol
mHCl = 1.8*36.5 = 65.7g
mdd HCl = 65.7*100/36=182.5g
VddHCl = mdd/D = 182.5/1.18=154.66ml