A 25.0 mL sample of a 1.20 M potassium sulfate solution is mixed with 14.6 mL of

Question

A 25.0 mL sample of a 1.20 M potassium sulfate solution is mixed with 14.6 mL of a 0.900 M barium nitrate solution and this precipitation reaction occurs: K2SO4(aq) + Ba(NO3)2(aq) rightarrow BaSO4(s) + 2KNO3(aq) The solid BaSO4 is collected, dried, and found to have a mass of 2.57 g. Determine the limiting reactant, the theoretical yield, and the percent yield. Determine the limiting reactant. Express your answer as a chemical formula. Determine the theoretical yield. Determine the percent yield.

Explanation / Answer

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

Moles of K2SO4 = Molarity * Volume ( in L) = 1.2 mol/L * 25 / 1000 L = 0.03 mol

Moles of Ba(NO3)2 = 0.900 mol/L * 14.6 / 1000 L = 0.013 mol

1 mol of K2SO4 requires 1 mol of Ba(NO3)2

0.03 mol of K2SO4 will require 0.03 mol of Ba(NO3)2

But mol of Ba(NO3)2 = 0.013 which is limited .

So, Ba(NO3)2 is the Limiting reagent

mol of BaSO4 = mol of Ba(NO3)2 = 0.013 mol

Mass of BaSO4 = 0.013 * Molar mass of BaSO4 = 0.013 mol * 233 g/mol = 3.06 g

Theoretical yied = 3.06 g

%yield = (Actual yield / Theoretical yield) * 100

%yield = ( 2.57 g / 3.06 g ) * 100 = 83.986%

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