Pretesst question just making sure I did it right ... chemistry helpers !!?

Question

Na2C2O4 + CaCl2 ~~~> CaC2O4 + 2NaCl

I put this together by this info

solid calcium oxalate formes by CaC2O4 and is used to give the calicum ion in a solution. How much in grams Na2C2O4 aka sodium oxalate is used to PPT the calcium ion from 37.5 mL of .104 M CaCl2 solution.

here is what I did

(37.5 mL) x (1 L/ 1000 mL) = .037 L

then I dont know is thats used but I did?

Na2C2O4 x (134 grams Na2C2O4 / 1 mol Na2C2O4 ) = 0.05226 Na2C2O4

is that the answer? or do I need the .037 L somewhere in there?

next part (2)

there is a mixture that contains 3.50 g He and 5.00 g Ne

calculate colume at STP

3.50 g He x (1 mol He)/ (4.002 g He) = .87456 mol He

5.00 g Ne x (1 mol Ne/20.18 g Ne) = .2477 g Ne

then do you add them and plug into PV=nRT?

.87456+.2477=1.12

so

PV=nRT

(1 atm)(V)=(1.12)(0.0821)(273.15)

is that right?

then it says calculate the partial pressure? which what is the equation for that and how do I plug in? I know I have this

Answer 1

Na2C2O4 + CaCl2 ~~~> CaC2O4 + 2NaCl

here is what I did

    (37.5 mL) x (1 L/ 1000 mL) = .037 L

That's ok.. but you didn't NEED to do it.  0.1 M is 0.1 moles/liter.. but it's ALSO 0.1 mmoles/mL.  You can just put in the 'milli's and go full speed ahead.

Try to stay out of the numbers as long as possible.  Think about the concepts and what you want to know.  According to the equation you wrote, the amount of oxalate you need is equal (in moles) to the amount of calcium chloride.  What's the amount of CaCl2?  That's was given to you that you have  37.5 mL of .104 Molar, or (37.5 * .104) millimoles.   That's about 4 millimoles, round numbers.  

That means you need 4 mmoles (ish) of oxalate.  You had the molecular weight of the oxalate (134 g/mole) but what you did with that wasn't clear.  Multiplying it by 4 millimoles, the amount you need, will give you the weight of that amount of oxalate: 134 mg/mmole *4 mmole = 500 mg, or 0.5 grams (again in very round numbers).  That could be what you did without writing it down, but if so one of us slipped a decimal.  That's how your volume got into the calculation (and concentration), through the calculation of the amount of calcium chloride that you were trying to equal.

Gasses:  you calculated the moles just fine.. but labeled the Ne as grams, not moles.  

And yes, once you know the total number of moles, you can just calculate the combined volume based on that number of moles, using PV=nRT.  

The PARTIAL pressure is simply the pressure due just to each gas.  So, it's PV=nRT, but using only the moles of the gas in question.  But, since you have the total pressure, the easy way is to realize that the partial pressures are equal to the mole fractions.. and just multiply the total pressure by (moles He)/[total moles]  to get partial pressure He.

Answer 2

First you need to compute the no of moles of CaCl2 in the 37.5 ml 0.104 M solution

No of moles CaCl2 = ( 0.375 L )( 0.104 mole / L )

                                  =  0.039 moles

From the given balance chemical equation for every mole of CaCl2 you need also one ( 1 ) mole of sodium oxalate so if you have 0.039 moles of CaCl2 you need 0.039 moles of sodium oxalate.

solving for the mass of sodium oxalate

grams of sodium oxalate = 0.039 moles X MWT of Na2C2O4

               = 0.039 moles x 134 grams / mole

                =  5.23 grams

therefore you need 5.23 grams of sodium oxalate

Part 2

Yes you are right V = 25.12 L

Partial pressure of He =( n of He ) * R * T  / V

Partial pressure of Ne = ( n of Ne )*R * T  / V

you just substitute the data in the given equation, hope this can help you