Question

For the following reaction, 3.29 grams of potassium sulfate are mixed with excess barium chloride. The reaction yields 3.91 grams of barium sulfate.

barium chloride (aq) + potassium sulfate (aq) -->barium sulfate (s) + potassium chloride (aq)
What is the theoretical yield of barium sulfate ? __ grams
What is the percent yield of barium sulfate ?

Answer 1

The theoretical yield of barium sulfate is calculated using stoichiometry from the known mass of potassium sulfate, and the percent yield is determined by comparing the actual yield to this theoretical yield.

Step-by-step explanation:

To calculate the theoretical yield of barium sulfate, we need the balanced chemical equation and the molar masses of the reactants and products.

The reaction is:

BaCl2 (aq) + K2SO4 (aq) → BaSO4 (s) + 2 KCl (aq)

First, calculate the moles of potassium sulfate (K2SO4) by dividing its mass by its molar mass. Then, use the stoichiometry of the balanced equation to find the moles of barium sulfate (BaSO4) that can be produced from the moles of potassium sulfate. Afterward, convert the moles of barium sulfate to grams using its molar mass to get the theoretical yield.

The percent yield can be found using the formula:

(Actual Yield / Theoretical Yield) x 100%

Where the actual yield is the mass of BaSO4 given (3.91 grams).

Answer 2

The theoretical yield of barium sulfate is approximately 4.41 grams, and the percent yield is approximately 88.66%.

To determine the theoretical yield of barium sulfate, we need to find the stoichiometric ratio between barium sulfate and potassium sulfate in the balanced chemical equation. The balanced equation is:

`$$\mathrm{BaCl}_2(a q)+\mathrm{K}_2 \mathrm{SO}_4(a q) \rightarrow \mathrm{BaSO}_4(s)+2 \mathrm{KCl}(a q)$$`

The molar ratio between barium sulfate
`$$\left(\mathrm{BaSO}_4\right)$$` and potassium sulfate
`$$\left(\mathrm{K}_2 \mathrm{SO}_4\right)$$` is 1:1. Therefore, the moles of barium sulfate formed should be equal to the moles of potassium sulfate reacted.

First, calculate the moles of potassium sulfate
`$$\left(\mathrm{K}_2 \mathrm{SO}_4\right)$$` using its molar mass:

`Molar \:mass \:of\: $\mathrm{K}_2 \mathrm{SO}_4=2 *$ atomic mass of $\mathrm{K}+$ atomic mass of $\mathrm{S}+4 *$ atomic mass of $\mathrm{O}$`

Now, calculate the moles of
`$$\left(\mathrm{K}_2 \mathrm{SO}_4\right)$$`:

`\text { Moles of } \mathrm{K}_2 \mathrm{SO}_4=\frac{\text { Mass of } \mathrm{K}_2 \mathrm{SO}_4}{\text { Molar mass of } \mathrm{K}_2 \mathrm{SO}_4}`

Next, since the molar ratio is 1:1 between
`$$\left(\mathrm{K}_2 \mathrm{SO}_4\right)$$` and
`\mathrm{BaSO}_4 \text {, }` the moles of
`$$\left(\mathrm{Ba} \mathrm{SO}_4\right)$$` formed are the same.

Now, calculate the theoretical yield of
`$$\left\mathrm{Ba} \mathrm{SO}_4\right$$` using its molar mass:

`\text { Theoretical yield }=\text { Moles of } \mathrm{BaSO}_4 * \text { Molar mass of } \mathrm{BaSO}_4`

Finally, calculate the percent yield using the formula:

`\text { Percent yield }=\left(\frac{\text { Actual yield }}{\text { Theoretical yield }}\right) * 100`

To calculate the theoretical yield and percent yield, we need to follow the steps outlined above.

Calculate moles of potassium sulfate
`$$\left(\mathrm{K}_2 \mathrm{SO}_4\right)$$`:

`Molar \:mass\: of\: $\mathrm{K}_2 \mathrm{SO}_4=(2 *$ atomic mass of $\mathrm{K})+($ atomic mass of $\mathrm{S})+$ $(4 *$ atomic mass of $\mathrm{O})$\\\\Molar mass of $\mathrm{K}_2 \mathrm{SO}_4=(2 * 39.10)+32.07+(4 * 16.00)$\\\\Molar mass of $\mathrm{K}_2 \mathrm{SO}_4=78.20+32.07+64.00$\\\\Molar mass of $\mathrm{K}_2 \mathrm{SO}_4=174.27 \mathrm{~g} / \mathrm{mol}$Moles of $\mathrm{K}_2 \mathrm{SO}_4=\frac{3.29 \mathrm{~g}}{174.27 \mathrm{~g} / \mathrm{mol}}$`

`\text { Moles of } \mathrm{K}_2 \mathrm{SO}_4 \approx 0.0189 \mathrm{~mol}`

Calculate theoretical yield of barium sulfate
`$$\left\mathrm{Ba} \mathrm{SO}_4\right$$`:

`The \:molar\: ratio\: between\: $\mathrm{K}_2 \mathrm{SO}_4$ and $\mathrm{BaSO}_4$ is $1: 1$.\\Theoretical yield $=$ Moles of $\mathrm{BaSO}_4 *$ Molar mass of $\mathrm{BaSO}_4$Theoretical yield $=0.0189 \mathrm{~mol} *(137.33+32.07+4 * 16.00)$\\Theoretical yield $\approx 0.0189 \mathrm{~mol} * 233.40 \mathrm{~g} / \mathrm{mol}$\\Theoretical yield $\approx 4.41 \mathrm{~g}$`

Calculate percent yield:

`Percent \:\:yield \:$=\left(\frac{\text { Actual yield }}{\text { Theoretical yield }}\right) * 100$\\Percent yield $=\left(\frac{3.91 \mathrm{~g}}{4.41 \mathrm{~g}}\right) * 100$\\Percent yield $\approx 88.66 \%$`

Question:

For the following reaction, 3.29 grams of potassium sulfate are mixed with excess barium chloride. The reaction yields 3.91 grams of barium sulfate.

barium chloride (aq) + potassium sulfate (aq) -->barium sulfate (s) + potassium chloride (aq)

What is the theoretical yield of barium sulfate ? __ grams

What is the percent yield of barium sulfate ?