Chemistry Problem-Please Help?

Question

1) How many grams of Na2Cr2O7 should be added to a 100.0-mL volumetric flask to prepare 0.025 M Na2Cr2O7 when the flask is filled w/ water?

2) You wish to prepare 0.12 M HNO3 from a stock solution of nitric acid that is 15.8 M. How many mL of the stock solution do you require to make up 1.00 L of 0.12 M HNO3?

3) A chemist added an excess of sodium sulfate to a solution of soluble barium compound to precipitate all of the barium ion as barium sulfate, BaSO4. How many grams of barium ion are in a 458-mg sample of the barium compound if a solution of the sample gave 513 mg BaSO4 precipitate? What is the mass % of barium in the compound?

I attempted these questions and my answer was way off from the right ones and I followed an example.

Answer 1

I'm guessing the errors were from having wrong units; it's very common.

For #1

Calculate moles of Na2Cr2O7 required:

100ml x 1l/1000ml 0.025mol/l = 0.0025 moles.

For grams to use, multiply moles x molecular weight of Na2Cr2O7:

0.0025 x [(2 x 23) + (2 x 52) + (7 x 16)]

= 0.0025mole x 262g/mole = 0.655g

for #2

Since you are making one liter, you can easily see you need 0.12 moles of HNO3; so you just need to find out what volume of the stock solution contains 0.12 moles of HNO3. (Watch units!)

0.12mol/15.8mol/l = 0.0076 Liters! or 7.6ml

For part #3:

All the barium that was in the unknown ended up in the BaSO4 that was recovered; so we need to find out how much barium is in 513mg of BaSO4.

First determine % barium by weight in BaSO4:

137.34/[137.34 + 32 + (4 x 16)] = 137.34/233.34 x 100% =

58.9%

Now find weight of barium in recovered BaSO4:

0.589 x 513mg BaSO4= 302mg Ba

Now find mass % Ba in soluble barium compound:

302mg/458mg x 100% = 65.9%