Respuesta :
The given question is incomplete. The complete question is as follows.
A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 212 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)
Explanation:
When [tex]Ba^{2+}[/tex] and [tex]Na_{2}SO_{4}[/tex] are added then white precipitate forms. And, reaction equation for this is as follows.
[tex]Ba^{2+} + SO^{2-}_{4} \rightarrow BaSO_{4}[/tex]
It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of [tex]BaSO_{4}[/tex] is 233.43.
Now, we will calculate the number of moles as follows.
No. of moles = mass × M.W
= [tex]\frac{0.212}{233.43}[/tex]
= 0.00091 mol of [tex]BaSO_{4}[/tex]
Hence, it means that 0.00091 mol of [tex]Ba^{2+}[/tex]. Now, we will calculate the mass as follows.
Mass = moles × MW
= [tex]0.00091 \times 137.327[/tex]
= 0.124 grams or 124 mg of barium
Thus, we can conclude that mass of barium into the original solution is 124 mg.
