Question

KBr contains 32.9 % potassium by mass. If 6.40g of $Br_2$ is made to react with 3.60 g of potassium the actual mass of potassium which reacted with $Br_2$ is

• A. 3.14 g
• B. l.76g
• C. 3.6 g
• D. None of these

Answer 1

Answer: (A)

3.14 g

Answer 2

$\%$ of bromine in $KBr=100-32.9=67.1\,\%$ $67.1\,g$ of bromine react with potassium $=32.9\,g$ $6.40\,g$ of bromine react with potassium $=\frac{\left(32.9\, g\right)\times\left(6.40\, g\right)}{\left(67.1 g\right)}$ $=3.138\,g$ $=3.14\,g$