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

Explanation

In KBr, 32.9 g of K are combined with 67.1 g of Br
`therefore" 3.6 g of K will combine with "Br_(2)=(67.1)/(32.9)xx3.6g=7.34" g which is not present"`
`"or 6.4 g or "Br_(2)" will combine with K"=(32.9)/(67.1)xx6.4=3.14g`
Thus, `Br_(2)` is the limiting reactant.
`"K reacted = 3.14 g"=(3.14)/(36)" mole = 0.08 mole"`
(K left unreacted `=(3.60-3.14)g=0.46g`)