Mixing the following solutions resulted in the isolation of 1.19 g of Ag 2 CrO 4

Question

Mixing the following solutions resulted in the isolation of 1.19 g of Ag2CrO4.

53.30 mL of 0.154 M AgNO3
53.50 mL of 0.118 M K2CrO4

Complete the following reaction table in millimoles

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

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

Explanation / Answer

The reaction will be

2AgNO3 + K2CrO4 --> Ag2CrO4 + 2KNO3

As per stoichiometry 2 moles of AgNO3 will react with 1 mole of K2CrO4 to give one mole of Ag2CrO4

Moles of Silver nitrate = molarity X Volume = 0.154 X 53.3 = 8.21 millimoles

Moles of K2CrO4 = 0.118 X 53.5 = 6.313 millimoles

Limiting reagent = Silver nitrate

so 8.21 millimoles of Silver nitrate will react with 4.105 millimoles of K2CrO4 to give 4.105 millimoles of Ag2CrO4

Moles of Ag2CrO4 = mass of Ag2CrO4 / mol wt of Ag2CrO4 = 1.19 / 331.7 = 0.00358 moles = 3.58 millimoles

What is the theoretical yield of Ag2CrO4 = 3.58 millimoles

What is the percent yield of Ag2CrO4 = 3.58 X 100 / 4.105 = 87.21 %

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---