Question

A 16.0 mL sample of a 1.56 M potassium sulfate solution is mixed with 14.3 mL of a 0.890 M barium nitrate solution and this precipitation reaction occurs: K2SO4(aq)+Ba(NO3)2(aq)→BaSO4(s)+2KNO3(aq) The solid BaSO4 is collected, dried, and found to have a mass of 2.51 g . Determine the limiting reactant, the theoretical yield, and the percent yield

Answer 1

Given,

Volume of Potassium Sulfate solution = 16 mL = 0.016 L

Molarity of potassium sulfate solution = 1.56 M

=> Moles of Potassium Sulfate = 0.016 x 1.56 = 0.02496 moles

Volume of Barium Nitrate solution = 14.3 mL = 0.0143 L

Molarity of Barium Nitrate solution = 0.89 M

=> Moles of Barium nitrate = 0.0143 x 0.89 = 0.012727 moles

K2SO4(aq)+Ba(NO3)2(aq)→BaSO4(s)+2KNO3(aq)

According to the stoichiometry of the above reaction, 1 mole of K2SO4 reacts with 1 mole of Ba(NO3)2

We have 0.02496 moles of K2SO4 and 0.012727 moles of Ba(NO3)2

Therefore Ba(NO3)2 is the limiting reagent

Theoretical Yield of BaSO4:

According to the stoichiometry of the above reaction,1 mole of Ba(NO3)2 reacts to produce 1 mole of BaSO4

=> Moles of BaSO4 produced = 0.012727 moles

Molar Mass of BaSO4 = 233.43

=> Mass of BaSO4 = 0.12727 x 233.43 = 2.97 g = Theoretical Yield of BaSO4

% yield = (2.51 / 2.97) x 100 = 84.5 %