Question

A 15.0 mL sample of a 1.74 M potassium sulfate solution is mixed with 15.0 mL of a 0.880 M barium nitrate solution, and a 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 2.48 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

Answer 1

The limiting reactant is potassium sulfate. The theoretical yield of barium sulfate is 6.09 g. The percent yield is 40.70%.

Step-by-step explanation:

To determine the limiting reactant, we need to calculate the number of moles of each reactant. Using the given concentrations and volumes, we can convert the volume to moles using the molarity formula. For the potassium sulfate solution, we have:

Moles of K2SO4 = 1.74 M x (15.0 mL / 1000 mL/L) = 0.0261 mol

For the barium nitrate solution, we have:

Moles of Ba(NO3)2 = 0.880 M x (15.0 mL / 1000 mL/L) = 0.0132 mol

Multiplying the moles of K2SO4 by the stoichiometric coefficient of BaSO4 (1), we find that the theoretical yield of BaSO4 is:

Theoretical yield = 0.0261 mol x (233.39 g/mol) = 6.09 g

To calculate the percent yield, we divide the actual yield (2.48 g) by the theoretical yield and multiply by 100:

Percent yield = (2.48 g / 6.09 g) x 100 = 40.70%