Al2O3 + 6HCl → 2AlCl3 + 3H2O
\(n_{Al_2O_3}=\frac{10,2}{102}=0,1\left(mol\right)\)
a) Theo PT: \(n_{HCl}=6n_{AlCl_3}=6\times0,1=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6\times36,5=21,9\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\frac{21,9}{400}\times100\%=5,475\%\)
b) \(m_{dd}saupư=10,2+400=410,2\left(g\right)\)
c) Theo pT: \(n_{AlCl_3}=2n_{Al_2O_3}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,2\times133,5=26,7\left(g\right)\)
\(\Rightarrow C\%_{AlCl_3}=\frac{26,7}{410,2}\times100\%=6,51\%\)
nAl2O3= 10.2/102=0.1 mol
Al2O3 + 6HCl --> 2AlCl3 + 3H2O
0.1______0.6______0.2
mHCl = 0.6*36.5=21.9g
C%HCl = 21.9*100/400=5.475%
mdd sau phản ứng = 10.2+400=410.2 g
mAlCl3 = 0.2*133.5=26.7 g
C%AlCl3 = 26.7/410.2*100%= 6.51%
