Answer: 43.3 grams
Explanation:-
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]
Thus [tex]\text{no of moles}of Na_2SO_4={0.0664M}\times {3.06 L}=0.203[/tex]
Thus [tex]\text{no of moles}of Ba(OH)_2={0.0820M}\times {2.27L}=0.186[/tex]
[tex]Ba(OH)_2(aq)+Na_2SO_4(aq)\rightarrow 2NaOH(aq)+BaSO_4(s)[/tex]
As 1 mole of [tex]Ba(OH)_2[/tex] reacts with 1 mole of [tex]Na_2SO_4[/tex]
0.186 moles of [tex]Ba(OH)_2[/tex] reacts with =[tex]\frac{1}{1}\times 0.186=0.186moles[/tex] of [tex]Na_2SO_4[/tex]
Thus [tex]Ba(OH)_2[/tex] is the limiting reagent and will limit the formation of the products.
As 1 mole of [tex]Ba(OH)_2[/tex] gives with 1 mole of [tex]BaSO_4[/tex] precipitate
0.186 moles of [tex]Ba(OH)_2[/tex] reacts with =[tex]\frac{1}{1}\times 0.186=0.186moles[/tex] of [tex]BaSO_4[/tex] precipitate
Mass of [tex]BaSO_4=moles\times {\text {Molar mass}}=0.186\times 233=43.3g[/tex]
Thus mass of the precipitate formed is 43.3 grams.
