Aluminum bromide can be prepared by reacting small pieces of aluminum foil with liquid bromine at room temperature. The balanced chemical reaction is:2Al(s) + 3Br2(l) → 2AlBr3(s)How many moles of Br2 are needed to produce 5 mol of AlBr3, if sufficient Al is present?

Respuesta :

Based on the mole ratio from the chemical reaction 3 moles of Br2 produces 2 moles of AlBr3. We can set this equation up to determine the unknown:

[tex]\begin{gathered} \frac{2}{3}=\frac{x}{5} \\ 3x=2\times5 \\ x=\frac{10}{3} \\ x=3.3moles \end{gathered}[/tex]

3.3 moles of Br2 is needed to reacts with 5 mol of AlBr3.