Barium sulfate was made by reacting 0.1188 L of a 0.4238 M potassium sulfate (K_

Question

Barium sulfate was made by reacting 0.1188 L of a 0.4238 M potassium sulfate (K_2SO_4) solution with 0.1411 L of a 0.3222 M barium nitrate (Ba(NO_3)_3) solution. The reaction of aqueous potassium sulfate and aqueous barium nitrate produces solid barium sulfate and aqueous potassium nitrate. Write the balanced chemical equation, an ionic equation, and a net ionic equation for the process Full equation: __ Complete ionic equation: __ Net ionic equation: __ Calculate the number of moles of potassium sulfate and barium nitrate. Which reactant is the limiting reagent? Determine the theoretical yield (in grams) of barium sulfate formed.

Explanation / Answer

Answer (a)

Full reaction :-

Ba(NO3)2(aq) + K2SO4(aq) --> BaSO4(s) + 2KNO3(aq)

Complete ionic equation :-

Ba2+ (aq) + 2NO3- (aq) + 2K+ (aq) + SO42- (aq) ---> BaSO4 (s) + 2K+ (aq) + 2NO3- (aq)

Net ionic equation :-

the nitrate ions and the sodium ions are in aqueous solution on both sides of the
equation, so they are spectator ions, leaving
Ba2+ (aq) + SO42- (aq) ---> BaSO4 (s)

Answer (b)

moles of K2SO4 = 0.1188L X 0.4238 mol /L = 0.0503 mole

moles of Barium Nitrate = 0.1411L X 0.3222 mol /L = 0.0454 mole

Answer (c)

Here moles of Barium Nitrate is less than moles of K2SO4 so Barium Nitrate is limiting reagent

Answer (d)

Here the molecular weight of Barium sulfate is 233.43 g/mol so

By considiring Barium Nitrate is limiting reagent and 100% theoretical yield

Barrium sulfate formed during process is = (233.43 g/mol) x (0.0454 mole) = 10.61 gm

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