## Chemistry 10th Edition

(a) $Na + I -- \gt NaI$ (b) $55.80g (NaI)$ are produced by the reaction of 47.24 g of iodine $(I)$.
(a) 1. Write and balance the reaction: $Na + I -- \gt NaI$ ** Since the exercise didn't give the formulas, we assume the reaction occurs between the sodium and iodine atoms. The reaction is already balanced. ------------- (b) 1. Calculate the number of moles of $I$: 126.90* 1 = 126.90g/mol $47.24g \times \frac{1 mol}{ 126.90g} = 0.3723mol (I)$ According to the balanced reaction: The ratio of $I$ to $NaI$ is 1 to 1: $0.3723 mol (I) \times \frac{ 1 mol(NaI)}{ 1 mol (I)} = 0.3723mol (NaI)$ 2. Calculate the mass of $NaI$: 22.99* 1 + 126.90* 1 = 149.89g/mol $0.3723 mol \times \frac{ 149.89 g}{ 1 mol} = 55.80g (NaI)$