Question

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)

Answer 1

12.3 moles of
`AlCl_(3)` are produced by reacting 18.4 moles of
`BaCl_(2)` .

Step-by-step explanation:

Balanced reaction:
`2AlPO_(4)+3BaCl_(2)\rightarrow 2AlCl_(3)+Ba_(3)(PO_(4))_(2)`

According to balanced equation:

3 molecules of
`BaCl_(2)` produce 2 molecules of
`AlCl_(3)`

So
`3N_(A)` molecules of
`BaCl_(2)` produce
`2N_(A)` molecules of
`AlCl_(3)` (
`N_(A)` is avogadro number).

We know 1 mol of molecule =
`N_(A)` number of molecules.

So 3 moles of
`BaCl_(2)` produce 2 moles of
`AlCl_(3)`.

Hence 18.4 moles of
`BaCl_(2)` produce
`((2)/(3)* 18.4)` moles of
`AlCl_(3)` or 12.3 moles of
`AlCl_(3)` .