Potassium bromide’ KBr contains 32.9% by mass of potassium. If 6.40 g of bromine reacts with 3.60 g of potassium, calculate the number of moles of potassium that combine with bromine to form KBr.
Mass percentage of K = 32.9 ∴ Mass percentage of Br = 100 – 32.9 = 67.1 Now 67.1 g of bromine combines with 32.9 g of potassium 6.40 g of bromine combines with = 32.967.1 × 6.40 = 3.14 g of K Potassium (K) which remains unreacted = 3.60 – 3.14 = 0.46 g Thus, Br2 is the limiting reagent. Mass of K = Mass of K Atomic mass of K=3.14 g39 g mol−1 = 0.0802 mol.
