Question

Determine the LIMITING reactant in the following balanced equation:

2KBr + CI2 --> 2KCI+ Br2, when 4g of KBr react with 6 g of CI . MM(KBr)=
119 g/mol MM(CI2)=70 g/mol *

Answer 1

KBr is limiting reactant.

Explanation:

Given data:

Mass of KBr =4g

Mass of Cl₂ = 6 g

Limiting reactant = ?

Solution:

Chemical equation:

2KBr + Cl₂ → 2KCl + Br₂

Number of moles of KBr:

Number of moles = mass/molar mass

Number of moles = 4 g/ 119 gmol

Number of moles = 0.03 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 6 g/ 70 gmol

Number of moles = 0.09 mol

Now we will compare the moles of reactant with product.

KBr : KCl

2 : 2

0.03 : 0.03

KBr : Br₂

2 : 1

0.03 : 1/2×0.03= 0.015

Cl₂ : KCl

1 : 2

0.09 : 2/1×0.09 = 0.18

Cl₂ : Br₂

1 : 1

0.09 : 0.09

Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.