Answer:
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 %
