Question

Answer 1

\(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)

\(n_{HCl}=0.15\cdot2=0.3\left(mol\right)\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(1.............6\)

\(0.1...........0.3\)

\(LTL:\dfrac{0.1}{1}>\dfrac{0.3}{6}\Rightarrow Al_2O_3dư\)

\(n_{AlCl_3}=0.1\left(mol\right)\)

\(m_{AlCl_3}=0.1\cdot133.5=13.35\left(g\right)\)

\(\Rightarrow D\)