Question
KBr (potassium bromide) 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 combines with bromine to form KBr.
Solution
As KBr contains 32·9% by weight of potassium, this means that 32·9 g of K combines with (100 – 32·9) i.e. 67·1 g of bromine.
67.1 g Br combines with 32.9 g K
6.40 g Br will combine with 
Thus, Br will be limiting reactant and K left unreacted
= (3.60 - 3.13) g
= 0.462 g
67.1 g Br combines with 32.9 g K 6.40 g Br will combine with Thus, Br will be limiting reactant and K left unreacted
= (3.60 - 3.13) g
= 0.462 g
