Question

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

Answer 1

Well, we interrogate the stoichiometric reaction:

`K(s) + 1/2Br_2(l) rarr KBr(s)`

Explanation:

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

`"Moles of potassium"=(2.45*g)/(39.10*g*mol^-1)=0.0626*mol`.

And thus the molar quantity of `0.0626` WITH RESPECT to `"potassium bromide"` is the same.........

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