[cladbolg]

Hey, I have tried everything possible to solve these questions, my book does not show us how to do them but gives them as exercise problems please help me out.

1) Calcium Nitrate will react with ammonium chloride at slightly elevated temperatures , as represented in the equation,
    Ca(NO3)2 (s) +2NH4Cl (s) ---> 2 N2O (g) + CaCl2 (s) + 4 H2O (g)
    A 3.00 mol sample of each reactant will give what volume of N2O at STP?

2) It takes 16.6 min for a 10.0-mL sample of an unknown gas to effuse through a pinhole. A 10.0-mL sample of helium He required 5.00min. What is the molecular weight of the unknown gas?

3) Which of the following gases, CF4, F2, H2, Ne, and SO3 has the slowest rate of diffusion under identical conditions? Explain teh answer.

for the second one,I tried solving it with the example on http://www.chemicalforums.com/index.php?board=2;action=display;threadid=2064;start=0 but how are we supposed to get the mass for the unknown gas with no moles given?

Please reply soon I have an exam on Tuesday morning.

[Borek]

Calcium Nitrate will react with ammonium chloride at slightly elevated temperatures , as represented in the equation,
    Ca(NO3)2 (s) +2NH4Cl (s) ---> 2 N2O (g) + CaCl2 (s) + 4 H2O (g)
    A 3.00 mol sample of each reactant will give what volume of N2O at STP?

Very simple stoichiometry. 1 mole of calcium nitrate gives 2 moles of N₂O, how many moles needed for 3 moles of N₂O? Use ratio/proportions to solve.

2) It takes 16.6 min for a 10.0-mL sample of an unknown gas to effuse through a pinhole. A 10.0-mL sample of helium He required 5.00min. What is the molecular weight of the unknown gas?

3) Which of the following gases, CF4, F2, H2, Ne, and SO3 has the slowest rate of diffusion under identical conditions? Explain teh answer.

for the second one,I tried solving it with the example on http://www.chemicalforums.com/index.php?board=2;action=display;threadid=2064;start=0 but how are we supposed to get the mass for the unknown gas with no moles given?

http://en.wikipedia.org/wiki/Graham's_law_of_effusion

There is even equation listed that you should use. You don't need any masses other then molar masses - and that for helium you should remember.

[cladbolg]

What i understand from the first question is that it is a limiting reagent question. 1 mol of
Ca(NO3)2 will give us 2 mol of N2O so 3 mol of Ca(NO3)2 will give us 6 mols of N2O, how do we find out volume now?

For the second and third how do I apply Graham's Law to it, like in the second I have min and mL given, to calculate rate will it just be 10.0mL/16.6min for the unknown and 10.0mL/5.00min for the He, divide them together for the rate ratio?

Basically will it be:
(10.0mL/16.6min)/(10.0mL/5.00min) = sqrt (4.00g/mol He/ mass unknown)? Then solve for mass unknown?

Please help me out.

[Borek]

What i understand from the first question is that it is a limiting reagent question.

Now that you wrote it looks to me like the question can be amiguous. I have read it "write how much N₂O will be produced from 3 moles calcium nitrate and how much from 3 moles ammonium chloride assuming there is an excess of the other reagent".

1 mol of
Ca(NO3)2 will give us 2 mol of N2O so 3 mol of Ca(NO3)2 will give us 6 mols of N2O, how do we find out volume now?

What is volume of one mol of any gas at STP?

(10.0mL/16.6min)/(10.0mL/5.00min) = sqrt (4.00g/mol He/ mass unknown)? Then solve for mass unknown?

Yep.

[cladbolg]

aha I never thought about checking for the volume of any gas at STP, I think that should help solve the first ya?

I can't believe second was that simple >.<

Any one know the answer to 3rd :O?

Thanks for the help, just a little more please

[Borek]

Any one know the answer to 3rd

Almost identical to 2^nd. Diffusion coefficient depends on molar mass.

[cladbolg]

so basically the higher the mass the slower the diffusion rate?

[Borek]

OTBE (Other Things Being Equal)

[cladbolg]

OTBE? I'm sorry I didn't understand that one could you please explain.

P.S. How do I get a scooby snack?

[Borek]

OTBE? I'm sorry I didn't understand that one could you please explain.

I already did

[cladbolg]

Here is what i came up with for number 2, STP, T=0C = 273K, and P = 1 atm. For 3 mol of each reactant we will get 3 mol of N2O since limiting reagent will be NH4CL. so, n=(PV)/(RT) where R is a constant = 0.0821 L*atm/(K*mol) and n is number of moles. Solve for V using that equation got 67.2 L. Seem correct?

Did second with formula posted above.

Still no clue about the third. OTBE (Other Things Being Equal) I am totally lost here.

Thanks for all the help Borek, could you also please help me with the last one  ?

About the Scooby Snacks >.>?

[Borek]

Here is what i came up with for number 2, STP, T=0C = 273K, and P = 1 atm. For 3 mol of each reactant we will get 3 mol of N2O since limiting reagent will be NH4CL. so, n=(PV)/(RT) where R is a constant = 0.0821 L*atm/(K*mol) and n is number of moles. Solve for V using that equation got 67.2 L. Seem correct?

Yes, but you did it hard way. It is enough to remember that 1 mole of gas at STP occupies 22.4L. 3*22.4 = 67.2L.

Still no clue about the third. OTBE (Other Things Being Equal) I am totally lost here.

OTBE means that if you are sure there are no other factors that can spoil the reasoning, you are right. For example: the higher engine power the faster the car is? OTBE true, if you take two cars with identical mass, shape and gears, the one with higher HP will be faster. But the 22 wheel truck is much slower then Porsche, even if it may have much stronger engine.

In the case of diffusion in gases you may assume thet other things are equal and molecule with higher molar mass diffuses slower. In the water solution that's not necessarilly true (although it is still true in many cases).

About the Scooby Snacks >.>?

You can be rewarded by others with scooby snacks. Help answering others questions and you will get some sooner or later.

[cladbolg]

  Thank you very much Borek I really appreciate the help I recieved. Wish there was more than a scooby snack I could give but hey I gave you a scooby snack  , Hmm wait there might be something else, Victoria's Secret models maybe? Thanks again, Clad.