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

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Answer 1 · 2017-04-24T16:41:14

Well, we interrogate the stoichiometric reaction:

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

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?

What molar quantity of "potassium bromide""potassium bromide" results from oxidation of a 2.45*g2.45*g mass of potassium?