[John_Kimble]

Hello everyone!

These are all simple exercises, but could someone please check over them quickly, just to be 100% sure, since there's a high possibility that they'll be used in the test.

I know it's quite a lot, but I will try to write down my procedures as clearly as I can. I'm confident they're mostly correct, but I'd really like to be sure.

Thanks a lot!!!



1) The following reaction produced 0.022 g of calcium oxalate (CaC2O4). What mass of calcium chloride was used as reactant?

CaCl2 (aq) + Na2C2O4 (aq) → CaC2O4 (s) + 2 NaCl (aq)
   
I calculate the # of moles of calcium oxalate ; then look at the stoichiometric equivalence between the 2 (1:1) and calculate the mass of CaCl2.

Result is 0.019063 g.


2) The principal component of many kidney stones is calcium oxalate. A kidney stone recovered from a typical patient contains 8.5 x 10^20 formula units of calcium oxalate. How many moles of calcium oxalate are present in this kidney stone? What is the mass of the kidney stone in grams?

a) I calculate the # of moles using Avogadro's # and the given formula units. Result is 0.0014115 mol.

b) Using the molar mass of the compound (128.1 g/mol) and result from a) I calculate the mass. Result is 0.1808 g.


3) Chloral hydrate, a sedative and sleep-inducing drug, is available as a solution labeled 648 mg/3.72 mL. What volume in mL should be administrated to a patient who is meant to recive 486 mg per dose?

I just use the simple crisscross calculation (sorry, don't know the right English term, I'm foreign ). The result is 2.79 mL.


4) Ferrous sulfate is one dietary supplement used to treat iron-deficiency anemia. How many milligrams of iron are in 250mg of ferrous sulfate?

I calculate the proportion of Fe in FeSO4 (it's 0.3676) then multiply it with 250 to get the amount of mgs in the given compound.


5) A 0.15 M solution of hydrochloric acid is used to titrate 30,0 mL of a calcium hydroxide solution of unknown concentration. If 140,0 mL of hydrochloric acid is required, what is the concentration (molarity) of the calcium hydroxide solution?

Ca(OH)2  +   2 HCl        CaCl2    +    2 H20

I calculate the # of moles of 0.15 M HCl solution in 30.0 mL (result is 0,021 mol). Then I look at the ratio (stoichiometric equivalence) between the 2 reagents (Ca(OH2) : HCl = 1:2). Therefore the amount of Ca(OH)2 needed is twice less than HCl.

Now I have the needed information to calculate the concentration of Ca(OH)2. Result is 0.35 M.

[Borek]

Your answers look good, but the last question is idiotic - Ca(OH)₂ doesn't have that high solubility.

[John_Kimble]

Thanks for the quick reply!

Yea, that's our school, they just make stuff up

[John_Kimble]

One more:

Calculate the molality of the phosphate acid solution having mass concentration of (this data is given, but I didn't manage to copy it in time, so I guess let's assume it's 1455 g/L - taken from http://www.csudh.edu/oliver/chemdata/acid-str.htm ). Density of the solution is 1.689 g/cm^3.

 
mass concentration= m (solute) / V (solution) = 1455 g/L

density= m (solute) + m (solvent) / V (solution) = 1689 g/L

Therefore, I subtract the mass concentration from the density to get the mass of the solvent, which is 234g. Mass of solute is then 1455g.

Now I calculate the # of moles of solute ( M(H3PO4= 98 g/mol), which is 14.847 mol.

So, the final result, the molality is: b= n / m (solvent) = 63.45 mol/kg.


Is this the right procedure, or did I mess it up somewhere? Thanks a lot!

[Borek]

Looks OK to me (but I just skimmed).

[John_Kimble]

I will trust your skimming abilities then Thank you.

It looks like a very easy / straightforward problem, I guess the "tricky part" is getting the mass of the solvent.