Question

KBr contains 32.9% by weight of potassium. If 6.40 g of bromine reacts with 3.60 g of potassium, calculate the number of moles of potassium which combine with bromine to form KBr.

Answer 1

For 100 gm of KBr sample, mass of K = 32.9 gm

Mass of Br = 100 -32.9 = 67.1 g

So, if 67.1 gm of bromine reacts with 32.9 gm of potassium

Then 6.40 gm Br will react with = (32.9/67.1) x 6.40 = 3.14 gm potassium.

So, in the given sample the amount of potassium which will be un reacted is = 3.60 -3.14 = 0.46 gm

Now, as we know that, mass of potassium reacting in the KBr sample = 3.14 gm

So, moles of potassium = given mass/ molar mass = 3.14/ 39 = 0.08 moles.

Hence, 0.08 moles of potassium combines to form KBr .