Stoichiometric problem... (1 Viewer)

[CHUDYMASTER]

I'm having a mental conflict here. Mclake made me believe him...but...

"Mass phosphate / mass barium phosphate = (94.97 * 2) / 601.9

= 189.94 / 601.84 = .3156 "

Could you explain that part?

 

[spice girl]

Originally posted by CHUDYMASTER
I'm having a mental conflict here. Mclake made me believe him...but...

"Mass phosphate / mass barium phosphate = (94.97 * 2) / 601.9

= 189.94 / 601.84 = .3156 "

Could you explain that part?

in barium phosphate, the percentage phosphate is constant.

it's equal to the mass of 2 units of phosphate, over the total formula mass of the barium phosphate.

so barium phosphate always has 31.56% phosphate.

so if you have 0.32g barium phosphate, you'll have 31.56% * 0.32g phosphate

 

[CHUDYMASTER]

You make a good point, thanks for clearing that up before I SUBMITTED MY ASSIGNMENT (mclake!! )

 

[CHUDYMASTER]

Actually, truth be told, I still don't see why McLake is wrong. And in that case, (because I did the rest the way he did it) could you tell me if these are correct?

Finding Sulphate Concentrations:

Ba2+ (aq) + SO42- (aq) ---> BaSO4 (s)

Mass of BaSO4 (s) = 1.42 g
Molar mass of BaSO4 (s) = 137.3 + 32.06 + (16 x 4) = 233.36 gmol-1
Moles of BaSO4 (s) = mass/molar mass = 1.42/ 233.36 = 6.09 x 10-3 mol

Since mol[BaSO4] = mol[SO42-],

Moles of SO42- = 6.09 x 10-3 mol
Molar mass of SO42- = 32.06 + (16 x 4) = 96.06 gmol-1
Mass of SO42- = moles x molar mass = 6.09 x 10-3 x 96.06 = 0.585 g

concentration (ppm or mg/L) = mass of sulphates (mg)/ volume of water (L):
Mass of SO42- = 585 mg
Volume of water = 0.1 L
Concentration of sulphate ions = 585/0.1 = 5850 ppm
Finding Chloride Concentrations:

Ag+ (aq) + Cl- (aq) --> AgCl (s)

Mass of AgCl (s) = 1.55 g
Molar mass of AgCl (s) = 107.9 + 35.45 = 143.35 gmol-1
Moles of AgCl (s) = mass/molar mass = 1.55/ 143.35 = 0.011 mol

Since mol[AgCl] = mol[Cl-],

Moles of Cl- = 0.011 mol
Molar mass of Cl- = 35.45 gmol-1
Mass of Cl- = moles x molar mass = 0.011 x 35.45 = 0.390 g

concentration (ppm or mg/L) = mass of chlorides (mg)/ volume of water (L):
Mass of Cl- = 390 mg
Volume of water = 0.1 L
Concentration of chloride ions = 390/ 0.1 = 3900 ppm

 

[McLake]

spice girl, arn't you calculating % of PO4 remaining?

(I think) My calculations calculate ppm of PO43- in the water, which is what the question asks.

 

[McLake]

Originally posted by CHUDYMASTER
Actually, truth be told, I still don't see why McLake is wrong. And in that case, (because I did the rest the way he did it) could you tell me if these are correct?

Without actually calculating values, these look correct (ie: the method looks right, if my previous mthod is right)

 

[CHUDYMASTER]

MCLAKE, I worked out our mistakes. You basically did the SAME THING as spice girl, BUT, there's a slight error in how you think about this part...

"now n[Ba3(PO4)2] = 2n[(PO4)3-]

so n[(PO4)3-] = 2.67 x 10 ^-4 "

NO!! You're thinking about it in terms of a mathematical equation. That first statement means that the phosphate ion IS 2 TIMES THE BARIUM PHOSPHATE IN AMOUNT.

In terms of mathematics, it should be read n = [Ba3(PO4)2]
2n = [(PO4)3-]. Now sub for n and get:

[(PO4)3-] = 2[Ba3(PO4)2]

*Sigh* Well at least I was right in the first place...

But it was an understandable mistake, thanks people.

 

[McLake]

Sorry Chuddy.