Jump to content
Sign in to follow this  
Catwalker

Chemistry II- This week: Gas Volume Problem

Recommended Posts

You have to notice that, even though its a 1-1 correlation between the number of OH- and H+ ions, you needed 4x the volume of the base to fully react the acid, so the acid must be 4x as concentrated. To actually work the problem out:

 

.04L * 1.0M NaOH = .04 moles NaOH = .04 moles OH- = .04 moles H+

 

.04 moles / .01L = 4.0M

Share this post


Link to post
Share on other sites

When you are talking about molarity of an acid, you are usually talking about the Hydrogen ion concentration. To figure out how many H+ ions there were in the 10 mL of acid, you need to figure out how many OH- ions it took to fully react the acid (since one H+ reacts with one OH-).

You have to notice that, even though its a 1-1 correlation between the number of OH- and H+ ions, you needed 4x the volume of the base to fully react the acid, so the acid must be 4x as concentrated. To actually work the problem out:

 

.04L * 1.0M NaOH = .04 moles NaOH = .04 moles OH- = .04 moles H+

 

.04 moles / .01L = 4.0M

The acid is unknown. What if it's diprotic? Hydrogen/hydroxyl concentration is normality, not molarity. 40mL of 1N NaOH would neutralize 10mL of 2-ish M H2SO4.

 

The question was probably confusing because there's no way to know the answer with the information given.

Share this post


Link to post
Share on other sites

Chemistry is so tedious, reminds me why I went into CS... *Goes and intersects two deterministic finite automatons using state cross products* :o

Share this post


Link to post
Share on other sites

Its a beginning Chemistry course, and probably the beginning of the acid/base part of that course, so I was working under the assumption it wasn't diprotic, because I doubt they want to worry about things like partial dissociation of the second proton.

 

And yes, titrations suck. I remember spending a good 2 or 3 weeks in HS Chem titrating potassium permanganate. That shit turns everything brown.

Share this post


Link to post
Share on other sites

You have to notice that, even though its a 1-1 correlation between the number of OH- and H+ ions, you needed 4x the volume of the base to fully react the acid, so the acid must be 4x as concentrated. To actually work the problem out:

 

.04L * 1.0M NaOH = .04 moles NaOH = .04 moles OH- = .04 moles H+

 

.04 moles / .01L = 4.0M

 

That was the answer I put because I kept coming up with .04 in my calculations.

 

What I'm going to do next time is work this at the same time I go through the chapter.

 

Chemistry is fun.

 

Except titration. Titration is boring.

 

I do enjoy the class. It's an online course, which I dislike for a class like this, but as long as I can devote enough time to it, it's not that bad. I'm not a big fan of math, really, so that part is challenging. But, I'm getting through it. I just have to pass it. I'd love to get an A or B, and I'm on that track so far.

 

I don't know how much titration I have to do. There was a small part of this last chapter that talked about it.

Share this post


Link to post
Share on other sites

The acid is unknown. What if it's diprotic? Hydrogen/hydroxyl concentration is normality, not molarity. 40mL of 1N NaOH would neutralize 10mL of 2-ish M H2SO4.

 

The question was probably confusing because there's no way to know the answer with the information given.

This is a simple titration problem to get him to use the equation M(acid) *V(acid) = M(Base) *V(Base)

 

: Molarity of acid X volume of acid = molarity of base * volume of base.

 

The thing to focus on here is that it says base neutralizes acid, IE moles H+ = OH-

 

 

EDIT: yeah question is bad. had it been say, H2SO4 like you said, the ratio would be 1:2, base:acid, so youd use 2M(acid) * V(acid) = M(base) *V(base). Diprotic acids are definitely in general chemistry. This is just a fail question.

 

 

Tell your instructor that his homework is wrong, and to replace your 80 with a 110.

 

EDIT 2: I guess it's best to say the formula is Na Va = Nb Vb, where N = normality, which is molarity taking into acct whether you have more than 1 moles of H+ or OH-. so N = molarity * (# of H+ or OH-)

 

Example: 10ml 1M Ca(OH)2 neutralizing 10ml portion of H3PO4: whats molarity of acid.

 

N acid = M*3.

N base = 1*2

 

3M *.010 = 2 *.010, M(acid) = .67.

Share this post


Link to post
Share on other sites

Okay, doing gasses now.

 

A sample of N2 gas occupies 2.40 L at 20oC. If the gas is in a container that can contract or expand at constant pressure, at what temperature will the N2 occupy 4.80 L?

a. 10 deg C

b. 40 deg C

c. 146 deg C

d. 313 deg C

e. 685 deg C

 

I'm fairly certain this falls under Charles's law, so since the volume of the gas has doubled, then it makes sense that the temperature of the gas has also doubled. So, the answer is B?

Share this post


Link to post
Share on other sites

Okay, doing gasses now.

 

A sample of N2 gas occupies 2.40 L at 20oC. If the gas is in a container that can contract or expand at constant pressure, at what temperature will the N2 occupy 4.80 L?

a. 10 deg C

b. 40 deg C

c. 146 deg C

d. 313 deg C

e. 685 deg C

 

I'm fairly certain this falls under Charles's law, so since the volume of the gas has doubled, then it makes sense that the temperature of the gas has also doubled. So, the answer is B?

 

Yes. It says constant pressure, so the gas contracts/expands with the same factor as the temp decreases/increases.

Share this post


Link to post
Share on other sites

Yes. It says constant pressure, so the gas contracts/expands with the same factor as the temp decreases/increases.

 

Cool. Thanks. I'm liking this chapter a little more.

 

Also, I found this site.

Share this post


Link to post
Share on other sites

This one seems easy, which has me worried that it's not so easy.

 

Which of the following gases will have the greatest density at the same specified temperature and pressure?

a. H2

b. CClF3

c. CO2

d. C2H6

e. CF4

 

All parts of the equation being equal (temperature, volume, pressure) the only variable would be the mass. So, if Density = mass/volume, the deciding factor would be the mass of the molecule. Since CF4 has the greatest molecular mass of all of them, I am thinking E.

Share this post


Link to post
Share on other sites

What mass of KClO3 must be decomposed to produce 126 L of oxygen gas at 133oC and 0.880 atm?

 

2KClO3(s) 2KCl(s) + 3O2(g)

a. 24.6 g

b. 70.8 g

c. 272 g

d. 408 g

e. 612 g

 

I am totally stuck on this problem. There are so many pre-calculations to make before I can just start on this one that by the time I think I'm close to the answer, none of what I come up with is related to any of the options.

 

Again, I don't need the answer, I just need to start somewhere. So, if you could help me with that....

Share this post


Link to post
Share on other sites

This one seems easy, which has me worried that it's not so easy.

 

Which of the following gases will have the greatest density at the same specified temperature and pressure?

a. H2

b. CClF3

c. CO2

d. C2H6

e. CF4

 

All parts of the equation being equal (temperature, volume, pressure) the only variable would be the mass. So, if Density = mass/volume, the deciding factor would be the mass of the molecule. Since CF4 has the greatest molecular mass of all of them, I am thinking E.

 

Right idea, but your mass calculations are wrong.

Share this post


Link to post
Share on other sites

Right idea, but your mass calculations are wrong.

 

I got:

H2= 2.016

CClF= 66.46

CO2= 44.01

C2H6=30.068

CF4=88.01

 

How are my mass calculations wrong? These are the molar masses of each molecule. If there's a different way to calculate mass of a molecule without knowing the exact amount, then I missed that in week one and I should just drop the class now.

Share this post


Link to post
Share on other sites

I got:

H2= 2.016

CClF= 66.46

CO2= 44.01

C2H6=30.068

CF4=88.01

 

How are my mass calculations wrong? These are the molar masses of each molecule. If there's a different way to calculate mass of a molecule without knowing the exact amount, then I missed that in week one and I should just drop the class now.

 

It's C-Cl-F-F-F, not C-Cl-F. You can double check some of this stuff when you have a bunch of electrons left over, then you probably screwed something really basic up.

Share this post


Link to post
Share on other sites

Damn. You're right. Simple math gets me a lot more often than anything else. Attention to detail.

 

Factoring in the CORRECT subscript, CClF3 has a molar mass of 104.51. And is the winner.

Share this post


Link to post
Share on other sites

What mass of KClO3 must be decomposed to produce 126 L of oxygen gas at 133oC and 0.880 atm?

 

2KClO3(s) 2KCl(s) + 3O2(g)

a. 24.6 g

b. 70.8 g

c. 272 g

d. 408 g

e. 612 g

 

I am totally stuck on this problem. There are so many pre-calculations to make before I can just start on this one that by the time I think I'm close to the answer, none of what I come up with is related to any of the options.

 

Again, I don't need the answer, I just need to start somewhere. So, if you could help me with that....

 

 

I screwed something up in my calculations and didn't get an answer, I'll work on it more after dinner.

 

But you are using PV = n RT. R being .0821 (L*atm), You can get the moles for oxygen, and then use that to find the total mass of the KClO3. You gave the chemical equation badly again, with no ->, and am i finding the mass of KClO3, or 2KClO3

Share this post


Link to post
Share on other sites

Grrr. I guess copy + paste doesn't bring over the symbols correctly.

 

It should be: 2KClO3(s) + 2KCl(s) --> 3O2(g)

 

I'll pay more attention next time. Sorry.

 

I'm guessing KClO3.

Share this post


Link to post
Share on other sites

I submitted my answers for this already. Don't worry about doing any more calculations, unless you just really want to.

Share this post


Link to post
Share on other sites

You get an answer? Im pretty sure I just made a stupid mistake or fail rounding along the way, the info I gave you should be correct.

 

Grrr. I guess copy + paste doesn't bring over the symbols correctly.

 

It should be: 2KClO3(s) + 2KCl(s) --> 3O2(g)

 

I'll pay more attention next time. Sorry.

 

I'm guessing KClO3.

 

Uh really? That arrow seems to be in the wrong spot.

 

EDIT: I did the calculation for you since you submitted already. I'll put it in spoilers. Issue was rounding error --- gogo fail rounding because using windows calc :P

 

 

Solving for 2KClO3(s) -> 2KCl(s) + 3O2(g)

 

PV = nRT

 

.88(atm) * 126(L) = n * .0821(L*atm) * 406K

 

n = 3.3265

 

n=m/M

3.3265 = m/96 <-- 3O2 = 32*3.

 

m = 319g.

 

formula weight of KClO3 is 122.551. O3 is 48. which is 39.167% O. so 319 is 39.167% of 815g. that's for 2KClO3, you just want mass of KClO3, so divide by 2 and you got 408g.

 

Share this post


Link to post
Share on other sites

Yeah, my fault. Arrow was in the wrong spot. Just transpose the + and the -->. That's what happens when I get tired. Sorry.

 

Okay, one more for this week and we should be done.

 

Calculate the heat of decomposition for this process at constant pressure and 25C:

 

CaCO3(s) --> CaO(s) + CO2(g)

 

(Symbols are in correct spot this time.)

 

According to a chart we have in our text

 

Enthalpy of CaCO3 = -1206.9

Enthalpy of CaO = -635.6

Enthalpy of CO2 = -393.5

 

Now, from how I understand this enthalpy thing, it's basically just addition and subtraction, like an algebraic equation and whatever I get from putting this all together in an equation is what should be the answer. Because it's in reverse, the energy should be positive.

 

-1206.9 --> -635.6 + -393.5

 

-1206.9 --> -1029.1

 

The difference between the two is -177.8 kj/mol. Because of the direction of the equation, the sign needs to be reversed, so the answer is 177.8 kj/mol.

Share this post


Link to post
Share on other sites
Now, from how I understand this enthalpy thing, it's basically just addition and subtraction, like an algebraic equation and whatever I get from putting this all together in an equation is what should be the answer. Because it's in reverse, the energy should be positive.

 

-1206.9 --> -635.6 + -393.5

 

-1206.9 --> -1029.1

 

The difference between the two is -177.8 kj/mol. Because of the direction of the equation, the sign needs to be reversed, so the answer is 177.8 kj/mol.

You are correct, but you should think about it in a different way.

 

ΔH = HFINAL-HINITIAL

 

ΔH = -1029.1 - (-1206.9) = 177.8 kj/mol

 

Remember the equation; if you start trying to figure out which way to flip signs, you will fuck it up much more often. You should also make sure to remember that a positive change in enthalpy means that heat was introduced to the system (endothermic) and a negative change means that heat was released from the system (exothermic).

Share this post


Link to post
Share on other sites

Thanks, boots. I'll check that out.

 

Here's a question that one of my classmates thinks is incorrect in how I worked it:

 

A .1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 3024 J/C. The temperature increases by 1.126C. Calculate the heat given off by the burning Mg, in kJ/g and kJ/mol.

 

Using the constant-volume calorimetry equation: qcal = (Ccal)(DeltaT)

 

qcal = 3.024kj/C * -1.126 = -3.41kj

 

.1375g Mg = .006 moles Mg

 

-3.41/.006 = -568.33 kj/mol

-568.33/24.31 = -23.38 kj/g

Share this post


Link to post
Share on other sites

Thanks, boots. I'll check that out.

 

Here's a question that one of my classmates thinks is incorrect in how I worked it:

 

A .1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 3024 J/C. The temperature increases by 1.126C. Calculate the heat given off by the burning Mg, in kJ/g and kJ/mol.

 

Using the constant-volume calorimetry equation: qcal = (Ccal)(DeltaT)

 

qcal = 3.024kj/C * -1.126 = -3.41kj

 

.1375g Mg = .006 moles Mg

 

-3.41/.006 = -568.33 kj/mol

-568.33/24.31 = -23.38 kj/g

your sig figs are wrong. You can't round stuff off, there are 4 sig figs for this. Answer is -24.76 kJ/g and -597.4 kJ/mol with the correct sig figs.

 

EDIT: I take that back, you did work it wrong. You just came to the right answer regardless.

 

qcal = 3.405 kJ. DT is Tfinal - Tinitial. It's positive.

 

qcal = -qrxn.

 

you are looking for the heat given off by burning Mg. So that's what makes it negative.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×