KBr contains 32,9 % potassium by mass. If 6.40 g of `Br_(2)` is made to react with 3.60 g of potassium, the actual mass of potassium which reacted with `Br_(2)` is

Options

A. 3.14 g
B. 1.76 g
C. 3.6 g
D. None of these

Answer

A

Explanation

% of bromine in KBr `= 100-32.9=67.1%`
67.1 g of bromin react with potassium = 32.9 g
6.40 g of bromine react with potassium
`=((32.9)xx(6.40g))/((67.2g)) = 3.138 g = 3.14 g`