Answer:
a) The theoretical yield is 408.45g of [tex]BaSO_{4}[/tex]
b) Percent yield = [tex]\frac{realyield}{408.45g}*100[/tex]
Explanation:
1. First determine the numer of moles of [tex]BaCl_{2}[/tex] and [tex]Na_{2}SO_{4}[/tex].
Molarity is expressed as:
M=[tex]\frac{molessolute}{Lsolution}[/tex]
- For the [tex]BaCl_{2}[/tex]
M=[tex]\frac{1.75molesBaCl_{2}}{1Lsolution}[/tex]
Therefore there are 1.75 moles of [tex]BaCl_{2}[/tex]
- For the [tex]Na_{2}SO_{4}[/tex]
M=[tex]\frac{2.0moles[tex]Na_{2}SO_{4}[/tex]}{1Lsolution}[/tex]
Therefore there are 2.0 moles of [tex]Na_{2}SO_{4}[/tex]
2. Write the balanced chemical equation for the synthesis of the barium white pigment, [tex]BaSO_{4}[/tex]:
[tex]BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl[/tex]
3. Determine the limiting reagent.
To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:
- For the [tex]BaCl_{2}[/tex]:
[tex]\frac{1.75}{1}=1.75[/tex]
- For the [tex]Na_{2}SO_{4}[/tex]:
[tex]\frac{2.0}{1}=2.0[/tex]
As the [tex]BaCl_{2}[/tex] 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.
[tex]1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}[/tex]
5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:
Percent yield = [tex]\frac{realyield}{theoreticalyield}*100[/tex]
Percent yield = [tex]\frac{realyield}{408.45g}*100[/tex]
The real yield is the quantity of barium white pigment you obtained in the laboratory.
