Mixing the following solutions resulted in the isolation of 3.97 g of Ag 2 CrO 4

Question

Mixing the following solutions resulted in the isolation of 3.97 g of Ag2CrO4.

87.20 mL of 0.316 M AgNO3
66.70 mL of 0.195 M K2CrO4

Complete the following reaction table in millimoles

A. What is the theoretical yield of Ag2CrO4?

B. What is the percent yield of Ag2CrO4?

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

D. Determine the equilibrium concentration of the limiting reactant.

2Ag1+ + CrO42- Ag2CrO4 Ksp = 1.1e-12 initial mmol delta mmol final mmol

Explanation / Answer

moles of AgNO3 used = 0.316 *87.27/1000 = 0.0275 moles

moles of K2CrO4 used = 0.195 *66.7/1000 = 0.013 moles

The reaction is

2AgNO3 + K2CrO4 -----> Ag2CrO4 + 2KNO3

2 moles of AgNO3 reacts with one mole of K2CrO4. 0.0275 moles AgNO3 reacts with 0.01375 moles of K2CrO4. But there is only 0.0130 mole K2CrO4. Thus K2CrO4 is the limiting reagent and it will be completely used. Amount of AgNO3 used = 2* 0.0130 = 0.026 moles

1 moles Ag2CrO4 is formed from 1 mole of K2CrO4

moles of Ag2CrO4 = 0.013 moles = 0.013 mole * 331.73 g/mole = 4.312 g

A. Theoretical yield = 4.312 g

B. percent yield = 3.97 *100/4.312 = 92.06 %

C.

total volume = 87.20 + 66.7 = 153.9 mL

comcentration = 0.0015/153.9 = 9.75 *10-6 M

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D.

Now, we have:
Ksp = 1.1e-12 = [Ag+]2[CrO4(2-)]
= (0.0776)^2*[CrO4(2-)]
Hence: [CrO4(2-)] = 1.1e-12/(9.75 *10-6)2 = 0.0105 (M)
This is the equilibrium concentration of the limiting reactant.

AgNO3 K2CrO4 Ag2CrO4 initial 0.0275 0.013 0 change -0.026 -0.013 +0.013 equilibrium 0.0015 x 0.013

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