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}=(50g)/(166g/mole)=0.301moles`

`\text{Moles of }Br_2=\frac{\text{Mass of }Br_2}{\text{Molar mass of }Br_2}=(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
`(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.