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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 and
are added then white precipitate forms. And, reaction equation for this is as follows.
It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of is 233.43.
Now, we will calculate the number of moles as follows.
No. of moles = mass × M.W
=
= 0.00091 mol of
Hence, it means that 0.00091 mol of . Now, we will calculate the mass as follows.
Mass = moles × MW
=
= 0.124 grams or 124 mg of barium
Thus, we can conclude that mass of barium into the original solution is 124 mg.
Answer:
Explanation:
Moles of SO₃ = 0.760 mol
Volume = 1.50 L
[SO₃] = 0.5067 M
Considering the ICE table for the equilibrium as:
Given:
Equilibrium concentration of O₂ = 0.130 mol
Volume = 1.50 L
[O₂] = x = 0.0867 M
[SO₂] = 2x = 0.1733 M
[SO₃] = 0.5067-2x = 0.3334 M
The expression for the equilibrium constant is: