\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{Al}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\\ \%m_{Al}=\dfrac{0,1.27}{7,8}.100\approx34,615\%\\ \Rightarrow\%m_{Al_2O_3}\approx65,385\%\\ Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ n_{Al_2O_3}=\dfrac{7,8-2,7}{102}=0,05\left(mol\right)\\ n_{HCl\left(t\text{ổ}ng\right)}=0,05.6+0,1.3=0,6\left(mol\right)\\ V_{\text{dd}HCl}=\dfrac{0,6}{3}=0,2\left(l\right)\)
\(a,n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{Al}=0,1\cdot27=2,7\left(g\right)\\ \Rightarrow\%_{Al}=\dfrac{2,7}{7,8}\cdot100\%\approx34,62\%\\ \Rightarrow\%_{Al_2O_3}\approx65,38\%\\ b,n_{Al_2O_3}=\dfrac{7,8-2,7}{102}=0,05\left(mol\right)\\ PTHH:Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ \Rightarrow n_{HCl}=2n_{H_2}+6n_{Al_2O_3}=0,6\left(mol\right)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,6}{3}=0,2\left(l\right)\)
Chỉ có al phản ứng với HCl sinh khí H2
2Al+6HCl-->2AlCl3 +3 H2
0.1 0 0.15
Al2O3+6HCl-->2AlCl3 +3 H2
0.05 0.3
nh2 = 3.36/22.4=0.15
nal=0.1 --> mal=27.0.1=2.7g
mal2o3=7.8-2.7=5.1g -> nal2o3=5.1/102=0.05
%mal2o3=5.1.100/7.8=65.385%
%mal=100-65.385=34.615%
Tổng mol HCl trong 2 pư là
0.3+0.3=0.6 mol
V=0.6/3=0.2l
