Answer
(a) $Na + I -- \gt NaI$
(b) $ 55.80g (NaI)$ are produced by the reaction of 47.24 g of iodine $(I)$.
Work Step by Step
(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.
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(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
$0.3723 mol \times \frac{ 149.89 g}{ 1 mol} = 55.80g (NaI)$
