Mixing the following solutions resulted in the isolation of 1.87 g of Ag2CrO4. 5
ID: 557011 • Letter: M
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
Explanation / Answer
59.50 mL of 0.337 M AgNO3 is: 20.05 mMol of AgNO3
55.40 mL of 0.115 M K2CrO4 is: 6.37 mMol of K2CrO4
Since 1 mole of CrO4(2-) may react with 2 moles of Ag+, 6.37mMol of K2CrO4 may react with 12.74 mMol of AgNO3.
Hence 6.37 mMol 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: 20.05mmol . . . . 6.37mmol . . . . 0.00mmol
delta: -12.74mmol . . . .-6.37mmol . . . . 6.37mmol
final: . 12.74mmol . . . . . . Xmmol . . . . . 6.37mmol
where X is NOT zero, since the initial and delta amounts of CrO4(2-) is NOT exactly the same and 6.37mmol is just an approximation with 3 significant figures. X is to be calculated in the following.
The total volume: 59.50mL + 55.40mL = 114.9 mL
The final [Ag+] is: (12.74mmol)/(114.9mL) = 0.1108M
This is the equilibrium concentration of the excess reactant.
Now, we have:
Ksp = 1.1e-12 = [Ag+]^2[CrO4(2-)]
= (0.1108)^2*[CrO4(2-)]
Hence: [CrO4(2-)] = 1.1*10^-12/(0.1108)^2 = 8.96*10^-11 (M)
This is the equilibrium concentration of the limiting reactant.
and thus X = (8.96*10^-11 M)*(114.9mL) = 1.03*10^-8 mmol
Since the molar mass of Ag2CrO4 is: 331.73 g/mol, the theoretical yield of Ag2CrO4 is:
(6.37mmol)*(331.73 g/mol) = 2.11 g
Thus the percent yield of Ag2CrO4 is: 1.87/2.11 = 0.886 = 88.6%