A15.0 ml sample of a 1.92 m potassium sulfate solution is mixed with 14.9 ml of a 0.860 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 2.46 g . determine the limiting reactant, the theoretical yield, and the percent yield. part a determine the limiting reactant. express your answer as a chemical formula.

The limiting reagent is barium nitrate, theoretical yield of barium sulfate is 3.03 g and percent yield of barium sulfate is 81.2 %

Explanation:

To calculate the number of moles for given molarity, we use the equation:

    .....(1)

Molarity of potassium sulfate = 1.92 M

Volume of solution = 15.0 mL = 0.015 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

Molarity of barium nitrate = 0.860 M

Volume of solution = 14.9 mL = 0.0149 L

Putting values in equation 1, we get:

For the given chemical reaction:

By Stoichiometry of the reaction:

1 mole of barium nitrate reacts with 1 mole of potassium sulfate

So, 0.013 moles of barium nitrate will react with = of potassium sulfate.

As, amount of potassium sulfate is more than the required amount. So, it is considered as an excess reagent.

Thus, barium nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of barium nitrate produces 1 mole of barium sulfate

So, 0.013 moles of barium nitrate will produce = of barium sulfate.

To calculate the mass of barium sulfate, we use the equation:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.013 moles

Putting values in equation 1, we get:

To calculate the percentage yield of barium sulfate, we use the equation:

Experimental yield of barium sulfate = 2.46 g

Theoretical yield of barium sulfate = 3.03 g

Putting values in above equation, we get:

Hence, the limiting reagent is barium nitrate, theoretical yield of barium sulfate is 3.03 g and percent yield of barium sulfate is 81.2 %

the ratio of the conjugate base to the acid required to attain the desired ph.