Question

Mixing the following solutions resulted in the isolation of 1.87 g of Ag2CrO4.

59.50 mL of 0.337 M AgNO3

55.40 mL of 0.115 M K2CrO4

Complete the following reaction table in millimoles 2Ag1+ + CrO42- Ag2CrO4 Ksp = 1.1e-12

initial mmol

delta mmol

final mmol

What is the theoretical yield of Ag2CrO4? g

What is the percent yield of Ag2CrO4? %

Assume additive volumes to determine the equilibrium concentration of the excess reactant. M

Determine the equilibrium concentration of the limiting reactant. M

Answer 1

Sol :-

Number of AgNO₃ = Molarity x Volume in L

= 0.337 mol/L x 0.05950 L

= 0.020 mol

= 20.05 mmol

Similarly,

Number of moles of K₂CrO₄ = 0.115 mol/L x 0.0554 L

= 0.006370 mol

= 6.37 mmol

Given reaction is :

2Ag⁺ (aq) + CrO₄^2- (aq) -------------> Ag₂CrO₄ (s)

Because, (given mole/stoichiometric coefficient) value of CrO₄^2- is smaller that is 6.37/1 = 6.37 than Ag⁺ which is 20.05/2 = 10.025, therefore CrO₄^2- will be the limiting reactant which is completely consumed in the reaction mixture.

Now, table is :

................................2Ag⁺ (aq)..............+..................CrO₄^2- (aq) ---------------------> Ag₂CrO₄ (s)

Initial........................20.05 mmol.................................6.37 mmol.......................................0.0 mol

Delta.........................-2x6.37 mmol............................-6.37 mmol........................................+6.37 mmol

Final..........................7.31 mmol......................................0.0 mmol.......................................6.37 mmol

Mass of Ag₂CrO₄ formed = Moles x Gram molar mass

= 6.37 x 10^-3 mol x 331.73 g/mol

= 2.11 g

Hence, theoretical yield of Ag₂CrO₄ = 2.11 g

Percent yield = Actual yield x 100% / Theoretical yield

= 1.87 g x 100% / 2.11 g

= 88.6%

Hence, percent yield of Ag₂CrO₄ = 88.6%

Total volume = 59.50 mL + 55.40 mL = 114.9 mL = 0.1149 L

So,

Equilibrium concentration of excessive reactant Ag⁺ = Moles/Volume in L

= 7.31 x 10^-3 mol / 0.1149 L

= 0.0636 mol/L

Hence, equilibrium concentration of excessive reactant Ag⁺ = 0.0636 M

Expression of K_sp of Ag₂CrO₄ is :

K_sp = [Ag⁺]²[CrO₄^2-]

[CrO₄^2-] = K_sp/[Ag⁺]²

= 1.1 x 10^-12 / (0.0636)²

= 2.72 x 10^-10 M

Hence, equilibrium concentration of limiting reactant CrO₄^2- = 2.72 x 10^-10 M