Answer
There are 1.27 moles of anions in 35.6 g of $AlF_3$.
Work Step by Step
1. Calculate the molar mass $(AlF_3)$:
26.98* 1 + 19.00 * 3 = 83.98g/mol
2. Calculate the number of moles $(AlF_3)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 35.6}{ 83.98}$
$n(moles) = 0.424$
3. Find the amount of anions:
- Each $AlF_3$ has 1 $Al^{3+}$ (Cation) and 3 $F^-$ (Anion)
Therefore, each mole of $AlF_3$ has 3 moles of anions.
$0.424 mol(AlF_3) \times \frac{3 mol(Anion)}{1 mol (AlF_3)} = 1.27 mol (Anion)$
