Question

Calculate the number of moles of excess reactant that will be left-over when 50.0 g of KI react with 50.0 g of Br2: 2KI + Br2 2KBr + I2

Answer 1

Hope this helps you.

Answer 2

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.