A 23.0 mL sample of a 1.12 M potassium sulfate solution is mixed with 14.3 mL of a 0.900 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 2.53 g . Determine the limiting reactant, the theoretical yield, and the percent yield

Question
Asked Nov 11, 2019
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A 23.0 mL sample of a 1.12 M potassium sulfate solution is mixed with 14.3 mL of a 0.900 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 2.53 g . Determine the limiting reactant, the theoretical yield, and the percent yield

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Expert Answer

Step 1

Number of mol of potassium sulfate = molarity x volume (L) = 1.12 M x 0.023 L = 0.02576 mol.

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