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

Answer 1

Wyatt Atkins

Well, we interrogate the stoichiometric reaction:

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

For this reason, the molar amount of potassium metal and potassium bromide are PRECISELY EQUIVALENT.

#"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.........

What would be the resulting mass of salt, given the quantitative yield?

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Answer 2

Cooper Adams

The molar quantity of potassium bromide formed from the oxidation of a 2.45 g mass of potassium is approximately 0.050 moles.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.