[coolman7992]

hey guys I am new here. Just wanted to say hi and also ask a few questions.

Right now I am currently doing some extra credit on questions relating to molarity which was a unit I did a few weeks ago. I am a little confused on this again, so if anybody could show me how to do these/give me the correct answer that would be much appreciated.

1. An alcoholic iodine solution ("tincture" of iodine) is prepared by dissolving 5.55 g of iodine crystals in enough alcohol to make a volume of 205 mL. Calculate the molarity of iodine in the solution.

for this one I know that there is the molarity equation but apparently .233 is wrong

2. Calcium oxalate, CaC2O4, is very insoluble in water. What mass of sodium oxalate, Na2C2O4 is required to precipitate the calcium ion from 35.5 mL of 0.101 M CaCl2 solution?

I was never taught how to do specific ions with molarity

3. What volume (in mL) of 0.15 M Na2SO4 solution is needed to precipitate all the barium, as BaSO4(s), from 12.5 mL of 0.45 M Ba(NO3)2 solution?

This stoich one is very confusing, this is the balanced equation but still confused where to go from here: Ba(NO3)2(aq) + Na2SO4(aq)  BaSO4(s) + 2 NaNO3(aq)

[Borek]

Please read forum rules.

[coolman7992]

is this better?

[Borek]

1. An alcoholic iodine solution ("tincture" of iodine) is prepared by dissolving 5.55 g of iodine crystals in enough alcohol to make a volume of 205 mL. Calculate the molarity of iodine in the solution.

for this one I know that there is the molarity equation but apparently .233 is wrong

Whait is formula of free iodine?

2. Calcium oxalate, CaC2O4, is very insoluble in water. What mass of sodium oxalate, Na2C2O4 is required to precipitate the calcium ion from 35.5 mL of 0.101 M CaCl2 solution?

I was never taught how to do specific ions with molarity

Start with the reaction equation, it is simple stoichiometry. Not harder then the next one, you just need to convert moles to mass.

3. What volume (in mL) of 0.15 M Na2SO4 solution is needed to precipitate all the barium, as BaSO4(s), from 12.5 mL of 0.45 M Ba(NO3)2 solution?

This stoich one is very confusing, this is the balanced equation but still confused where to go from here: Ba(NO3)2(aq) + Na2SO4(aq)  BaSO4(s) + 2 NaNO3(aq)

No idea what confuses you - you need 1 mole of Na₂SO₄ per each mole of Ba(NO₃)₂. How many moles of Ba(NO₃)₂ do you have?

[coolman7992]

ohhhh Iodine is diatomic. So if I were to solve the problem would the answer .106 be correct?

and for number 2, what kind of reaction is it? Im not sure about that one, but once i got the reaction do i find the moles in CaCl2 then use that in a stoich problem with Ca? (really confused)

and number 3 do I find out the mL of Na2SO4, then ocne I find that find the moles of both Na2SO4 and Ba(NO3)2. After I find the moles i convert to grams then do a stoich equation with Barium on the product side? (also confused)

[Borek]

ohhhh Iodine is diatomic. So if I were to solve the problem would the answer .106 be correct?

OK

and for number 2, what kind of reaction is it?

Almost identical to that from the third question. You mix two solutions and you get a precipitate.

Im not sure about that one, but once i got the reaction do i find the moles in CaCl2 then use that in a stoich problem with Ca? (really confused)

Yes.

and number 3 do I find out the mL of Na2SO4, then ocne I find that find the moles of both Na2SO4 and Ba(NO3)2. After I find the moles i convert to grams then do a stoich equation with Barium on the product side? (also confused)

Answer the question I have posted - how many moles of Ba(NO₃)₂ do you have? Then remember that you need 1 mole of sodium sulfate to precipitate out barium from 1 mole of barium nitrate.