Question

A 15.0 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.

Answer 1

Answer: 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:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$     .....(1)

• For potassium sulfate:

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:

$1.92M=\frac{\text{Moles of potassium sulfate}}{0.015L}\\\\\text{Moles of potassium sulfate}=(1.92mol/L\times 0.015L)=0.029mol$

• For barium nitrate:

Molarity of barium nitrate = 0.860 M

Volume of solution = 14.9 mL = 0.0149 L

Putting values in equation 1, we get:

$0.860M=\frac{\text{Moles of barium nitrate}}{0.0149L}\\\\\text{Moles of barium nitrate}=(0.860mol/L\times 0.0149L)=0.013mol$

For the given chemical reaction:

$K_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2KNO_3(aq.)$

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 = $\frac{1}{1}\times 0.013=0.013mol$ 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 = $\frac{1}{1}\times 0.013=0.013mol$ of barium sulfate.

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.013 moles

Putting values in equation 1, we get:

$0.013mol=\frac{\text{Mass of barium sulfate}}{233.4g/mol}\\\\\text{Mass of barium sulfate}=(0.013mol\times 233.4g/mol)=3.03g$

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

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of barium sulfate = 2.46 g

Theoretical yield of barium sulfate = 3.03 g

Putting values in above equation, we get:

$\%\text{ yield of barium sulfate}=\frac{2.46g}{3.03g}\times 100\\\\\% \text{yield of barium sulfate}=81.2\%$

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 %