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

KBr (potassium bromide) contains 32.9% by weight of potassium. If 6.40

1.	KBr (potassium bromide) contains 32.9% by weight of potassium. If 6.40 g of bromine react with 3.60 g of potassium, calculate the number of moles of potassium which combine with bromine to form KBr.
Answer» Solution :The AMOUNT of potassium in KBr = 32.9% This MEANS that 67.1 parts of bromine combine with 32.9 parts of potassium to form KBr. `therefore`The amount of potassium that reacts with 6.40 g of bromine `=(32.9)/(67.1) XX 6.40 = 3.14 g` `therefore` The number of moles of potassium PRESENT in this amount = `3.14/39.10 = 8.03 xx 10^(-2)` Hence, `8.03 xx 10^(-2)`moles of potassium combine to form KBr.