# Chapter 3 - Mass Relationships in Chemical Reactions - Section Problems - Page 106: 49

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. $0.424 mol(AlF_3) \times \frac{3 mol(Anion)}{1 mol (AlF_3)} = 1.27 mol (Anion)$

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