Question

34.62 g of ferrous bromide reacts with 29.25 g of potassium chloride in a heated vessel. Write a balanced chemical equation for this double replacement reaction.  What is the percent yield of ferrous chloride if the actual yield is 10.25 g?  How much excess remains?  

Answer 1

In given reaction, Reactants are 1) Ferrous Bromide ( FeBr ₂ ) 2 ) potassium chloride ( KCl ) and products are 1) Ferrous chloride (  FeCl ₂ ) and 2) potassium bromide ( KBr )

Molar mass of FeBr ₂ = 55.85 + 2 ( 79.91 ) = 215.67 g / mol

Molar mass of KCl = 39.10 + 35.45 = 74.55 g / mol

Molar mass of FeCl ₂ = 55.85 + 2 ( 35.45 ) = 126.75 g / mol

Molar mass of KBr = 39.10 + 79.91 = 119.01 g / mol

Balanced chemical equation for double displacement reaction is , FeBr ₂ + 2 KCl FeCl ₂ + 2 KBr

From reaction, 1 mol FeBr ₂ 2 mol KCl

i e 215.67 g FeBr ₂ 2 74.55 g KCl

34.62 g FeBr ₂ ( 2 74.55 34.62 / 215.67 ) g KCl

34.62 g FeBr ₂ 23.93 g KCl

For complete reaction of 34.62 g ferrous bromide , 23.93 g KCl is required. Provided amount of KCl is 29.25 g .

Provided amount of KCl is more than required amount therefore KCl is excess reactant and  FeBr ₂ is limiting reactant.

Now, calculate theoretical yield of ferrous chloride.

From reaction, 1 mol FeBr ₂  1 mol FeCl ₂

i e 215.67 g FeBr ₂ 126.75 g FeCl ₂

34.62 g FeBr ₂( 126.75 34.62 / 215.67 ) g FeCl ₂

34.62 g FeBr ₂ 20.35 g FeCl ₂

Theoretical yield of ferrous chloride is 20.35 g.

We know that, % yield of product = ( Actual yield of product / Theoretical yield of product ) 100

% yield of ferrous chloride = 10.25 g / 20.35 g ) 100

= 50.37 %

Excess reactant is KCl.

Amount of KCl remained after reaction = Initial amount of KCl - amount of KCl consumed in the reaction.

= 29.25 g - 23.93 g

= 5.32 g