Question

Potassium bromide (KBr) contains 32.9% by mass of potassium. Calculate the mass of bromine which combines with 3.6 g of potassium to form potassium bromide.

Answer 1

Mass of K= 32.9
Mass of Br= total - K=  100 - 32.9= 67.1
for the reaction of 6.4 g of Br with K= 32.9/67.1 x 6.4
=3.14 g

The amount of un reacted K= 3.6-3.14= 0.46 g

moles of K= givn mass/ molar mass
= 3.14/ 39 
=0.08moles 

Answer 2

Answer:

Explanation:

Suppose mass of KBr = 100g

Mass of Potassium = 32.9g

Mass of bromine = 100 - 32.9 = 67.1g

67.1g of bromine will combine with 32.9g of potassium

So 6.4g of bromine will react with 32.9/67.1 x 6.4 = 3.14g

Atomic mass of potassium = 39u

Required number of moles = 3.14/39 = 0.0805moles

Hope it helps