Mixing the following solutions resulted in the isolation of 1.54 g of Ag 2 CrO 4
ID: 855745 • Letter: M
Question
Mixing the following solutions resulted in the isolation of 1.54 g of Ag2CrO4.
88.10 mL of 0.327 M AgNO3
30.30 mL of 0.170 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 Mixing the following solutions resulted in the isolation of 1.54 g of Ag2CrO4. 88.10 mL of 0.327 M AgNO3 30.30 mL of 0.170 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. MExplanation / Answer
88.10 mL of 0.327 M AgNO3 is: 28.8mMol of AgNO3
30.30 mL of 0.170 M K2CrO4 is: 5.15 mMol of K2CrO4
Since 1 mole of CrO4(2-) may react with 2 moles of Ag+, 5.15mMol of K2CrO4 may react with 14.4mMol of AgNO3. Hence 5.15mMol of K2CrO4 is the limiting reactant.
Since Ksp is very small, almost all K2CrO4 got reacted to form Ag2CrO4(s). Thus we can provide:
. . . . . . . Ag+ . . . . . . . . CrO4(2-) . . . . . Ag2CrO4
initial: 28.8mmol . . . . 5.15mmol . . . . 0.00mmol
delta: -14.4mmol . . . .-5.15mmol . . . . 5.15mmol
final: . 14.4mmol . . . . . . Xmmol . . . . . 5.15mmol
where X is NOT zero, since the initial and delta amounts of CrO4(2-) is NOT exactly the same and 1.15mmol is just an approximation with 3 significant figures. X is to be calculated in the following.
The total volume: 88.10mL + 30.30mL = 118.40mL
The final [Ag+] is: (14.40mmol)/(118.40mL) = 0.122 M
This is the equilibrium concentration of the excess reactant.
Now, we have:
Ksp = 1.1e-12 = [Ag+]^2[CrO4(2-)]
= (0.122)^2*[CrO4(2-)]
Hence: [CrO4(2-)] = 1.1e-12/(0.0776)^2 = 0.74e-10 (M)
This is the equilibrium concentration of the limiting reactant.
and thus X = (0.74e-10 M)*(118.40mL) = 0.88e-8 mmol
Since the molar mass of Ag2CrO4 is: 331.73 g/mol, the theoretical yield of Ag2CrO4 is:
(5.15mmol)*(331.73 g/mol) = 1.71 g
Thus the percent yield of Ag2CrO4 is: 1.54/1.71 = 90.06%