Question

ni. Ja What is the percentage yield of silver chromate if you mix solutions containing 0.019300 mols of potassium chromate with 0.00920 mols of silver nitrate, and actually produce 0.9886 g of silver chromate? 2AgNO3(aq) + K2 CrO4(aq) - Ag2 CrO4(s) + 2KNO3(aq)

Answer 1

Balanced chemical equation is:
2 AgNO3 + K2CrO4 ---> Ag2CrO4 + 2 KNO3


2 mol of AgNO3 reacts with 1 mol of K2CrO4
for 9.2*10^-3 mol of AgNO3, 4.6*10^-3 mol of K2CrO4 is required
But we have 1.93*10^-2 mol of K2CrO4

so, AgNO3 is limiting reagent
we will use AgNO3 in further calculation


Molar mass of Ag2CrO4,
MM = 2*MM(Ag) + 1*MM(Cr) + 4*MM(O)
= 2*107.9 + 1*52.0 + 4*16.0
= 331.8 g/mol

According to balanced equation
mol of Ag2CrO4 formed = (1/2)* moles of AgNO3
= (1/2)*9.2*10^-3
= 4.6*10^-3 mol


use:
mass of Ag2CrO4 = number of mol * molar mass
= 4.6*10^-3*3.318*10^2
= 1.526 g

% yield = actual mass*100/theoretical mass
= 0.9886*100/1.526
= 64.77%
Answer: 64.77%