Stuck on Chem Questions

[luvblueyes]

Hi...I have to add a few more questions to the list....
I feel embarrassed b/c some of these are REALLY simple, but I just would like them explained please

Question 43.
For the reaction b/w A and B to form C it is found that when one combines 0.6 moles of A with 0.6 moles of B, all of the B reacts, 0.2 moles of A reamin UNREACTED and 0.4 moles of C are produced. What is the balanced equation for this reaction?

The answer is 2A + 3B --> 2C

Can someone just explain? I figured out the 3B and 2C, but I wasn't sure how to know about the 2A part if that makes any sense.


Question 56.
For the reaction:
AgCl(s) + 2NH3(aq) --><-- Ag(NH3)2+(aq) + Cl-(aq)
the equilibrium constant K = 4x10^-3, which of the following statements is true?
[Ksp for AGCl is 1.0 X10^-10]

A. The addition of NH3 decreases the solubility of AgCl.
B. AgCl is more soluble in aqueous NH3 than in water.
C. AgCl is more soluble in aqueous solution containing Cl- than water.
D. AgCl is less soluble in aqueous NH3 than in water.

The answer is B

Please explain


Question 58.
What would be the heat of formation, &#916;H, for NO2gas if one considers the equations for the following reactions where all substances are gases?
1/2N2 + 1/2O2 --> NO Hf=+21.6kcal
NO2 --> NO + 1/2O2 H= +13.5kcal

A. -28.7 kcal
B. -8.1 kcal
C. 35.1 kcal
D. 28.7 kcal
E. 8.1 kcal

The answer is E.

Sorry...this one might not make sense unless you look at the actual question.




Anyways, my brain isn't working properly and I can't figure these out! If anybody can offer some help, it would be very very appreciated!!!!!!

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[OnionKnight]

Hey there,

For question 43, it says 0.6 moles of A are there to begin with, and then 0.2 moles remain unreacted. That means you had 0.4 moles of A reacted, producing 0.4 moles of C. Due to stoichiometry, A and C have to be the same ratio. In other words, you had 0.4 moles of A + 0.6 moles of B --> 0.4 moles C. That will give you 2A + 3B --> 2C

I'm not sure how familiar you are with Ksp, but this is the solubility product. It's kind of a way to tell you how soluble something is in water. You see that Ksp < K. When you write out an expression for K, it's generally concentration of products divided by that of reactants. If you have a larger K, that means products are more favored (smaller K means reactants more favored). Since K in the reaction is larger than Ksp, that means AgCl dissolving is more favored, which means AgCl dissolves better in ammonia.

Number 58 is kind of difficult to tell without the actual question. Hopefully those explanations up there make sense. Well, I hope this helps in your studying!

 

[luvblueyes]

Hey there,

For question 43, it says 0.6 moles of A are there to begin with, and then 0.2 moles remain unreacted. That means you had 0.4 moles of A reacted, producing 0.4 moles of C. Due to stoichiometry, A and C have to be the same ratio. In other words, you had 0.4 moles of A + 0.6 moles of B --> 0.4 moles C. That will give you 2A + 3B --> 2C

I'm not sure how familiar you are with Ksp, but this is the solubility product. It's kind of a way to tell you how soluble something is in water. You see that Ksp < K. When you write out an expression for K, it's generally concentration of products divided by that of reactants. If you have a larger K, that means products are more favored (smaller K means reactants more favored). Since K in the reaction is larger than Ksp, that means AgCl dissolving is more favored, which means AgCl dissolves better in ammonia.

Number 58 is kind of difficult to tell without the actual question. Hopefully those explanations up there make sense. Well, I hope this helps in your studying!

Yes!!! It helped so much! I'm just letting myself get overwhelmed. I'm a little high strung at the moment I am taking the OAT on thursday morning so when I find something I don't immediately understand, I get all flustered and panick!!

Thanks again!!!!

 

[ToePUG]

For Question 58 you want to look at which reactions will form NO2.

1/2N2 + 1/2O2 --> NO Hf=+21.6kcal
NO2 --> NO + 1/2O2 H= +13.5kcal

From these equations we see that the second one has NO2, but it's not being formed, it being broken down. So to form NO2 we must reverse the sign of H.

NO + 1/2O2 --> NO2 H= -13.5kcal

Now from this equation we can see that NO takes part in the reaction which itself requires 21.6kcal to form (from the first equation).

Therefore you will have to put in 21.6kcal for the first reaction, and 13.5kcal will be released in the second reaction. 21.6 - 13.5 = 8.1

Good luck Thursday!

 

[luvblueyes]

For Question 58 you want to look at which reactions will form NO2.

1/2N2 + 1/2O2 --> NO Hf=+21.6kcal
NO2 --> NO + 1/2O2 H= +13.5kcal

From these equations we see that the second one has NO2, but it's not being formed, it being broken down. So to form NO2 we must reverse the sign of H.

NO + 1/2O2 --> NO2 H= -13.5kcal

Now from this equation we can see that NO takes part in the reaction which itself requires 21.6kcal to form (from the first equation).

Therefore you will have to put in 21.6kcal for the first reaction, and 13.5kcal will be released in the second reaction. 21.6 - 13.5 = 8.1

Good luck Thursday!

I swear, this is the EXACT way I tried it Haha, well it didn't work out for me, but I tried it again and apparantly it does!!

Thank you so much for your help!!!