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

1 Answer
Apr 24, 2017

Answer:

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?