Few questions

BH20

Question 3: Yep, you are right. It is 100degreescelsius. I was looking at a lot of things at once, the book, the screen, and some webpages, going back and forth to understand the steps, and looked at the pressure instead which was 110.

Question 4: I cant say I quiet understand the ratio proportion stuff...I understand why you have to do it, but I dont quiet get it. I understood what you did in the other question, but not here.

Do I have to get all the MW of all the compounds in the equation ? They are as follows...
KClO3 is 121.55g, KCL is 74g and O2 is 48g

I get that since 3/2 is infront of the O2, when finding the MW, you have to times it by 1.5..but then, I dont get it after that.. because after you did this proportion stuff wouldn't you already have a mass and be done the question after 3 steps?

Gokul43201

The equation says that for every mole of KClO3 used, you get 1.5 moles of water.

Think of it like say, a cooking recipe : "1 cup of sugar goes into making 4 cups of cake. How much sugar is contained in half a cup of cake, if a cup of sugar weighs 160gm ?"

4 cups cake => 1 cup sugar, so 0.5 cups cake => 0.5*1/4 = 0.125 cups sugar => 0.125*160 = 20 gm.

It's the same kind of thing with the equations. Replace 'cup' with 'mole' and 'weight of a cup' with the 'MW'.

Now using this in Q4 :

n(O2) = 0.195 moles from PV = nRT for oxygen.
But 1.5 moles(O2)=>1 mole(KClO3) So, 0.195 mol(O2)=>0.195*1/1.5 = 0.13 mol(KClO3) =>0.13*121.55g=15.8g of KClO3

NOTE : The MW is simply the weight of 1 mole of the compound (like the weight of 1 cup of sugar)...not the weight of 'so many moles'. So the MW of O2 is always 32g. Then you can multiply this number by the number of moles of O2 to fing the weight of O2 - though we don't need that for this problem.

Hope this helps.

BH20

Ok, let me try to explain this out loud to see if it makes any sense...

- We are trying to find the mass of pottassium chlorate (KClO3).
- We are given the info for the oxygen gas (O2), in which we will be able to find the number of moles. So we use the Ideal Gas Law(PV=nRT). We found it to be 0.195moles. And we want to find moles because along with the MW, we can then find the mass of KClO3.

- We then have to look at the ratio proportion..O2 is 1.5moles, and KCL03 is 1 mole.
To find it we use the moles we found O2(0.195*1(KClO3)/1.5...we get 0.13moles....

- Now we can use the moles of KClO3 (0.13moles) and times it by its MW (121.55) to get its mass of 15.8g. (using formula n=m/MW)

- Ok, now I realize all the steps, and how we came to the answer. I'm pretty sure though it will take another couple of questions to really get it going, because its much easier when you actually see it done.

One thing that confused me was the we found O2 to be 0.195moles, but I also saw 1.5moles infront of it, so I was getting confused. BUT, what I gather it means is that O2 is 0.195moles, but so its propertionate and balanced, it will take 1.5moles for the whole equation to be balanced..is that correct?

I'll get to question 5 and give it a try, but for it...it says "Determine the quantity in moles of the gas that remains unreacted". so how to we know which one that is?

Gokul43201

The equation for the reaction only tells you the ratio of how many mole react with how many moles to give how many moles etc. If we multiply the equation throughout by some contant number, it is still correct. In a way, that is exactly what we are doing.

KClO3 --> KCl + 1.5 O2
can also be written as

5KClO3--> 5KCl + 7.5 O2

or 100KClO3--> 100KCl + 150 O2

For simplicity, it is written in the first form or sometimes multiplied by 2 so there are no fractions. What we are doing is multiplying this equation by some number which then makes the coefficient of O2 to become 0.195. Clearly that number is 0.195/1.5 = 0.13

So then you can write :

[tex]\frac {0.195} {1.5} KClO_3 --> \frac {0.195} {1.5} KCl + \frac {0.195} {1.5} *1.5 O_2 [/tex]

OR...


[tex]0.13 KClO_3 --> 0.13 KCl + 0.195 O_2 [/tex]

So this tells you how many moles of KClO3 were used up.

Like you said, familiarity leads to comprehension. Do more problems and you'll get the hang of it.

Gokul43201

For Q5, convert the volumes into moles using the Gas Law. Now find the constant that you need to multiply the reaction equation by to give you this number of moles - there will be 2 numbers, one that works for NH3 and another that works for O2 - the smaller of these 2 numbers is the right one. Think about why it is. So this will tell you how many moles get used up (of NH3 and O2 - one of them, completely) in the reaction. But you know how many moles were put into the container. So the difference is the number of unreacted moles.

BH20

sadly, I'm really confused about this WHOLE question, even though it is abit like the other one. Not sure what you even mean by some of the things. Never converted volume into moles before....I dont actually know what to do here at all.

Gokul43201

You'll understand once you start.

You're kidding ?! How did you find the no. of moles of O2 in Q4 from the volume of O2 ?

BH20

So we will use the Ideal Gas Law, and put in both volumes and covert them into moles..and then the right answer will be the lower moles one? or I should say the one we will use?

PV=nRT
n(0.20m3)=PV/RT
n=(800atm)(0.20m3)/(0.0821)(523K)
n=160/42.9393
n=3.726moles

n(0.12m3)=2.235moles

hmm..this doesn't look right..I've tried some other similar questions, and I didn't get them either.

I really apoligize, but I seriously dont get this one...

Gokul43201

First make sure you understand what units to use when. If you use R=0.0821 L-atm/K-mol, then
V : liters,
P : atm
T : K

If you use R = 8.315 J/K-mol
V : m^3
P : Pa or N/m^2
T : K

Now what values of n(O2) and n(NH3) do you get ?

BH20

Yeah, I used the 0.821 (R) so I knew to use L, atm and K as well, but the number was too high.

because it was 800atm*200L? or is that right?

Gokul43201

It will be high @ 800 atm, so that's okay.

BH20

PV=nRT
n(02)=PV/RT
n=(800atm)(200L)/(0.0821)(523K)
n=160,000/42.9383
n(O2)=3,726.63moles

n(NH3)=2,235.77moles

ok, so I get N(O2) 3,726.63 moles. n(NH3) is 2,235.77moles.

Gokul43201

I think it's the other way round.

BH20

yep, sorry.

n(NH3) 3,726.63 moles. n(O2) is 2,235.77moles.

Gokul43201

Yes, now if you multiply the equation

NH3(g) + 1.75 O2(g) -> NO2(g) + 1.5 H2O(g) (just changed fractions to decimals)
throughout by the number 2235.77/1.75=1277.6 , you get

1277.6 NH3(g) + 2235.77 O2(g) -> 1277.6 NO2(g) + (1.5*1277.6) H2O(g)

This converts the equation to a form that involves the actual number of moles in the reaction vessel. So, this says that all the O2 will get used up but only 1277.6 moles of NH3 will react. So the unreacted number is 3726.6 - 1277.6 = 2449 moles of NH3.

BH20

alright thanks..can we get to the the writing equations and balancing? Maybe that is the reason I did not pick this up abit better.

sinc sulphate + barium nitrate produces ................
gold (III) sulphate + barium chloride produces.......
tin + antimony(V) chloride produces...
ferrous bromide + phosporic acid produces....
the complete combustion of ethene, C2H4 produces...

when writing equations, do you basically look at the ions? I do know though how the compounds come together.

Gokul43201

You need to read up the different types of reactions...I cannot describe them all here. And you must understand valency, oxidation numbers, etc. Look up your elementary chemistry textbook.

Also see : http://members.tripod.com/~EppE/reaction.htm

Start at : http://members.tripod.com/~EppE/index.htm

Start from the intro and go through stoichiometry (and Gas Laws, if necessary)

PS : these pages take a while to load...thanks to a slow ad-server.