Question

.02 mol of aluminum is burned in an excess of oxygen and the product is reacted with 2.0 mol/dm3 of HCl.
what minimum volume of acid will be required for complete reaction?

Answer 1

\[\Huge 4Al+3O_2 \rightarrow 2Al_2O_3\]

Answer 2

Then
\[\LARGE Al_2O_3+6HCl \rightarrow 2AlCl_3 + 3H_2O\]

Answer 3

To get moles of \(Al_2O_3\),
\[n(Al_2O_3)=0.02molAl \times \frac{2molAl_2O_3}{4molAl}=0.01molAl_2O_3\]

Answer 4

Then we find the number of moles of HCl, \(n(HCl)\) by using \(n(Al_2O_3)\)
\[n(HCl)=0.01molAl_2O_3 \times \frac{6molHCl}{1molAl_2O_3}=0.06molHCl\]

Answer 5

Using
\[[HCl]=\frac{n(HCl)}{V}\]
\[V=\frac{n(HCl)}{[HCl]}\]
\[V=\frac{0.06}{2.0}=0.03dm^3\]

Answer 6

Thank you so much.