Answer : The moles of excess reactant, \(Br_2\) is, 0.1625 mole.
Explanation : Given,
Mass of \(KI\) = 50 g
Mass of \(Br_2\) = 50 g
Molar mass of \(KI\) = 166 g/mole
Molar mass of \(Br_2\) = 160 g/mole
First we have to calculate the moles of \(KI\) and \(Br_2\).
\(\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}=\frac{50g}{166g/mole}=0.301moles\)
\(\text{Moles of }Br_2=\frac{\text{Mass of }Br_2}{\text{Molar mass of }Br_2}=\frac{50g}{160g/mole}=0.313moles\)
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
\(2KI+Br_2\rightarrow 2KBr+I_2\)
From the balanced reaction we conclude that
As, 2 moles of \(KI\) react with 1 mole of \(Br_2\)
So, 0.301 moles of \(KI\) react with \(\frac{0.301}{2}=0.1505\) moles of \(Br_2\)
From this we conclude that, \(Br_2\) is an excess reagent because the given moles are greater than the required moles and \(KI\) is a limiting reagent and it limits the formation of product.
The moles of excess reactant, \(Br_2\) = Given moles - Required moles
The moles of excess reactant, \(Br_2\) = 0.313 - 0.1505 = 0.1625 mole
Therefore, the moles of excess reactant, \(Br_2\) is, 0.1625 mole.
