2 AIPO4 + 3 BaCl2 --> 2 AICIz + 1 Baz(PO4)2

how many moles of aluminum chloride are produced by reacting 18.4 moles of barium chloride? (don't forget your units)

Respuesta :

Answer:

12.3 moles of [tex]AlCl_{3}[/tex] are produced by reacting 18.4 moles of [tex]BaCl_{2}[/tex] .

Explanation:

Balanced reaction: [tex]2AlPO_{4}+3BaCl_{2}\rightarrow 2AlCl_{3}+Ba_{3}(PO_{4})_{2}[/tex]

According to balanced equation:

3 molecules of [tex]BaCl_{2}[/tex] produce 2 molecules of [tex]AlCl_{3}[/tex]

So [tex]3N_{A}[/tex] molecules of [tex]BaCl_{2}[/tex] produce [tex]2N_{A}[/tex]  molecules of [tex]AlCl_{3}[/tex] ([tex]N_{A}[/tex] is avogadro number).

We know 1 mol of molecule = [tex]N_{A}[/tex] number of molecules.

So 3 moles of [tex]BaCl_{2}[/tex] produce 2 moles of [tex]AlCl_{3}[/tex].

Hence 18.4 moles of [tex]BaCl_{2}[/tex] produce [tex](\frac{2}{3}\times 18.4)[/tex] moles of [tex]AlCl_{3}[/tex] or 12.3 moles of [tex]AlCl_{3}[/tex] .