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 found to have a mass of 258 mg. What mass of baium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction)

Na2SO4(aq) + Ba(aq)-> BaSO4(s) + 2Na(aq)

.258gr BaSO4 x 1 mol BaSO4/ 233.4 gr BaSO4 x 1 mol Ba/1 mol BaSO4 x 137.33 grams Ba/1 mol Ba = .152 grams of Ba

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 266 mg.

Answer: m(Ba+2)=157 mg

Balanced Chemical Equation: Ba + Na2SO4 --> BaSO4 + 2 Na

Well, well, well, looks like someone stirred up a precipitate puzzle! Let's do some chemical clowning to find the answer!

First things first, we need to figure out what mass of barium sulfate (BaSO4 ) was formed. Given that the mass of the precipitate is 258 mg, it means we have 258 mg of BaSO4.

Now, let's do some clown chemistry! The molar mass of BaSO4 is 233.38 g/mol.

So, if we have 258 mg of BaSO4, we can use the molar mass to find the moles of BaSO4.
258 mg = 0.258 g.

Now, to find the moles, we divide the mass by the molar mass of BaSO4:
0.258 g / 233.38 g/mol ≈ 0.001105 mol.

Since barium sulfate has a 1:1 ratio with barium ions, we know that 0.001105 mol of BaSO4 equates to the same number of barium moles.

Now, let's do the final circus trick! The molar mass of barium (Ba) is 137.33 g/mol.

To find the mass of barium, we multiply the number of moles by the molar mass of barium:
0.001105 mol x 137.33 g/mol ≈ 0.1517 g.

So, the mass of barium in the original solution is approximately 0.1517 grams. Ta-da! The clown chemistry has spoken!

To determine the mass of barium in the original solution, we need to use stoichiometry and the mass of the precipitate.

First, let's write the balanced chemical equation for the reaction:

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

From the equation, we can see that 1 mole of barium ions (Ba²⁺) reacts with 1 mole of sulfate ions (SO₄²⁻) to form 1 mole of barium sulfate (BaSO₄). Therefore, the molar ratio between barium and barium sulfate is 1:1.

To calculate the moles of barium sulfate, we first need to convert the mass of the precipitate from milligrams to grams:

Mass of precipitate (BaSO₄) = 258 mg = 0.258 g

Next, we need to determine the molar mass of barium sulfate (BaSO₄):
- Molar mass of Ba = 137.33 g/mol
- Molar mass of S = 32.07 g/mol
- Molar mass of O = 16.00 g/mol (x 4 since there are 4 oxygen atoms in the formula)

Molar mass of BaSO₄ = (137.33 g/mol) + (32.07 g/mol) + (16.00 g/mol x 4) = 233.39 g/mol

Now we can calculate the moles of barium sulfate using the formula:

Moles of BaSO₄ = Mass of BaSO₄ / Molar mass of BaSO₄
= 0.258 g / 233.39 g/mol

Finally, since the molar ratio between barium and barium sulfate is 1:1, we know that the moles of barium ions are equal to the moles of barium sulfate:

Moles of Ba²⁺ = Moles of BaSO₄

To find the mass of barium (Ba) in the original solution, we multiply the moles of barium ions by the molar mass of barium:

Mass of Ba = Moles of Ba²⁺ x Molar mass of Ba

Now, you can substitute the values to calculate the mass of barium in the original solution.

258 mg BaSO4 = 0.258 g.

mols BaSO4 = 0.258/molar mass BaSO4.
mols Ba = same as mols BaSO4,
g Ba = mols x atomic mass Ba