Question

If a 32.4 gram sample of sodium sulfate (Na2SO4) reacts with a 65.3 gram sample of barium chloride (BaCl2) according to the reaction below: Na2SO4 (aq) + BaCl2 (aq) → BaSO4 (s) + 2NaCl (aq) What is the theoretical yield of barium sulfate (BaSO4) in grams?

Answer 1

Answer: The theoretical yield of barium sulfate is 50.9 grams

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For sodium sulfate:

Given mass of sodium sulfate = 32.4 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

`\text{Moles of sodium sulfate}=(32.4g)/(142g/mol)=0.228mol`

• For barium chloride:

Given mass of barium chloride = 65.3 g

Molar mass of barium chloride = 208.23 g/mol

Putting values in equation 1, we get:

`\text{Moles of barium chloride}=(65.3g)/(208.23g/mol)=0.314mol`

The chemical equation for the reaction of barium chloride and sodium sulfate follows:

`Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl`

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of barium chloride

So, 0.228 moles of sodium sulfate will react with =
`(1)/(1)* 0.228=0.228mol` of barium chloride

As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

1 mole of sodium sulfate produces 1 mole of barium sulfate.

So, 0.228 moles of sodium sulfate will produce =
`(1)/(1)* 0.228=0.228moles` of barium sulfate

Now, calculating the mass of barium sulfate from equation 1, we get:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.228 moles

Putting values in equation 1, we get:

`0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol* 223.4g/mol)=50.9g`

Hence, the theoretical yield of barium sulfate is 50.9 grams