A 22.0 mLmL sample of a 1.16 MM potassium sulfate solution is mixed with 14.8 mLmL of a 0.860 MM barium nitrate solution and this precipitation reaction occurs: K2SO4 (aq) + Ba (NO3) 2 (aq) →BaSO4 (s) + 2KNO3 (aq) K2SO4 (aq) + Ba (NO3) 2 (aq) →BaSO4 (s) + 2KNO3 (aq) The solid BaSO4BaSO4 is collected, dried, and found to have a mass of 2.57 gg. Determine the limiting reactant, the theoretical yield, and the percent yield.

The limiting reactant is Ba (NO3) 2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %

Explanation:

Step 1: Data given

Volume of a 1.16 M potassium sulfate solution (K2SO4) = 22.0 mL = 0.022 L

Volume of a 0.860 M barium nitrate solution (Ba (NO3) 2 = 14.8 mL = 0.0148 L

The solid BaSO4 is collected, dried, and found to have a mass of 2.57 grams

Step 2: The balanced equation

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

Step 3: Calculate moles

Moles = volume * molarity

Moles K2SO4 = 0.022 L * 1.16 M

Moles K2SO4 = 0.02552 moles

Moles Ba (NO3) 2 = 0.0148 L * 0.860 M

Moles Ba (NO3) 2 = 0.012728 moles

Step 4: Calculate the limiting reactant

For 1 mol K2SO4 we need 1 mol Ba (NO3) 2 to produce 1 mol BaSO4 and 2 moles KNO3

Ba (NO3) 2 is the limiting reactant. It will completely be consumed. (0.012728 moles). K2SO4 is in excess. There will remain 0.02552 - 0.012728 = 0.012792 moles

Step 5: Calculate moles BaSO4

‬For 1 mol K2SO4 we need 1 mol Ba (NO3) 2 to produce 1 mol BaSO4 and 2 moles KNO3

For 0.012728 moles Ba (NO3) 2 we'll have 0.012728 moles BaSO4

Step 6: Calculate mass BaSO4

Mass BasO4 = moles BaSO4 * molar mass BaSO4

Mass BaSO4 = 0.012728 moles * 233.38 g/mol

Mass BaSO4 = 2.97 grams

Step 7: Calculate the percent yield

% yield = (actual yield / theoretical yield) * 100 %

% yield = (2.57 grams / 2.97 grams) * 100 %

% yield = 86.5 %

The limiting reactant is Ba (NO3) 2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %