What molar quantity of "potassium bromide" results from oxidation of a 2.45*g mass of potassium?

Apr 24, 2017

Well, we interrogate the stoichiometric reaction:

$K \left(s\right) + \frac{1}{2} B {r}_{2} \left(l\right) \rightarrow K B r \left(s\right)$

Explanation:

And thus the molar quantity of potassium bromide is PRECISELY EQUIVALENT to the molar quantity of potassium metal.........

$\text{Moles of potassium} = \frac{2.45 \cdot g}{39.10 \cdot g \cdot m o {l}^{-} 1} = 0.0626 \cdot m o l$.

And thus the molar quantity of $0.0626$ WITH RESPECT to $\text{potassium bromide}$ is the same.........

Given quantitative yield, what mass of the salt would result?