Question

The concentration of a saturated BaCl2 solution is 1.75 M (mol/liter) and the concentration of a saturated Na2SO4 solution is 2.0 M. Calculate the theoretical yield, in grams, and the percent yield for your synthesis of the barium white pigment.

Answer 1

a) The theoretical yield is 408.45g of
`BaSO_(4)`

b) Percent yield =
`(realyield)/(408.45g)*100`

Step-by-step explanation:

1. First determine the numer of moles of
`BaCl_(2)` and
`Na_(2)SO_(4)`.

Molarity is expressed as:

M=
`(molessolute)/(Lsolution)`

- For the
`BaCl_(2)`

M=
`(1.75molesBaCl_(2))/(1Lsolution)`

Therefore there are 1.75 moles of
`BaCl_(2)`

- For the
`Na_(2)SO_(4)`

M=
}{1Lsolution}[/tex]

Therefore there are 2.0 moles of
`Na_(2)SO_(4)`

2. Write the balanced chemical equation for the synthesis of the barium white pigment,
`BaSO_(4)`:

`BaCl_(2)+Na_(2)SO_(4)=BaSO_(4)+2NaCl`

3. Determine the limiting reagent.

To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:

- For the
`BaCl_(2)`:

`(1.75)/(1)=1.75`

- For the
`Na_(2)SO_(4)`:

`(2.0)/(1)=2.0`

As the
`BaCl_(2)` is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

`1.75molesBaCl_(2)*(1molBaSO_(4))/(1molBaCl_(2))*(233.4gBaSO_(4))/(1molBaSO_(4))=408.45gBaSO_(4)`

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield =
`(realyield)/(theoreticalyield)*100`

Percent yield =
`(realyield)/(408.45g)*100`

The real yield is the quantity of barium white pigment you obtained in the laboratory.