Bài 1:
a) Al2O3 + 6HCl → 2AlCl3 + 3H2O
\(n_{Al_2O_3}=\dfrac{5,1}{102}=0,05\left(mol\right)\)
\(m_{HCl}=200\times3,65\%=7,3\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
Theo PT: \(n_{Al_2O_3}=\dfrac{1}{6}n_{HCl}\)
Theo bài: \(n_{Al_2O_3}=\dfrac{1}{4}n_{HCl}\)
Vì \(\dfrac{1}{4}>\dfrac{1}{6}\) ⇒ Al2O3 dư
b) Các chất sau phản ứng: AlCl3, H2O và Al2O3 dư
Theo PT: \(n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=\dfrac{1}{3}\times0,2=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=\dfrac{1}{15}\times133,5=8,9\left(g\right)\)
Theo PT: \(n_{H_2O}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times0,2=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2O}=0,1\times18=1,8\left(g\right)\)
Theo PT: \(n_{Al_2O_3}pư=\dfrac{1}{6}n_{HCl}=\dfrac{1}{6}\times0,2=\dfrac{1}{30}\left(mol\right)\)
\(\Rightarrow n_{Al_2O_3}dư=0,05-\dfrac{1}{30}=\dfrac{1}{60}\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=\dfrac{1}{60}\times102=1,7\left(g\right)\)
\(\Sigma m_{chất}saupư=m_{AlCl_3}+m_{H_2O}+m_{Al_2O_3}dư=8,9+1,8+1,7=12,4\left(g\right)\)
\(\Rightarrow\%AlCl_3=\dfrac{8,9}{12,4}\times100\%=71,77\%\)
\(\%H_2O=\dfrac{1,8}{12,4}\times100\%=14,52\%\)
\(\Rightarrow\%Al_2O_3dư=100\%-14,52\%-71,77\%=13,71\%\)
