Question

Mixing the following solutions resulted in the isolation of 2.22 g of Ag2CrO4. 81.70 mL of 0.207 M AgNO3 51.60 mL of 0.144 M K2CrO4 Complete the following reaction table in millimoles 2Ag1+ + CrO42- Ag2CrO4 Ksp = 1.1e-12

initial Correct: Your answer is correct. Correct: Your answer is correct. Correct: Your answer is correct. mmol

delta Incorrect: Your answer is incorrect. Correct: Your answer is correct. Correct: Your answer is correct.

mmol final Incorrect: Your answer is incorrect. Correct: Your answer is correct. Correct: Your answer is correct. 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

Number of mill moles of AgNO3 = 81.70 * 0.207 = 16.9119 m moles

Number of mill moles of K2CrO4 = 51.60 * 0.1440 = 7.4304 m moles

The reaction will be

2Ag+ + CrO4(2-) -------> Ag2CrO4

Initial 16.92 7.43 0

Delta (16.92-2*7.43) 0 7.43

Final 2.06 0 7.43

Molar mass of Ag2CrO4 = 2(107.86) + 1(51.99) + 4(15.9996) = 331.74 gm/mol

Theoritical Yield = number of moles * molar mass

=> 7.43 * 10^(-3) * 331.74

=> 2.464 grams

Percent Yield = actual Yield/Theoritical Yield * 100

=> 2.22/2.464 * 100 = 90.09%

Final concentration of Ag+ = 2.06/(81.70+51.60) = 0.0154M

Ksp = 1.1 * 10^(-12) = [Ag+]^2[CrO4(2-)]

1.1 * 10^(-12) = (0.0154)^2 * [CrO4(2-)]

[CrO4(2-)] = 4.638 * 10^(-9) M