YT2095
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basic math help :(
ok for a start I suck at math! I get an Instant mental block and then even stuff I DO KNOW vanishes also 
a simple question in college today:
Q: What mass of KClO3 must be heated to give 1 litre of gas at STP using the equation:
2KClO3 ---> 2KCl + 3O2
===============================
now I figured that 2x03 is the same as 3xO2 (6 singlet oxygens each), so my Ratio was 1:1
apparently I borked it up and it should have been a 2:3 ratio as stated in the RXN formula.
I just don`t get it at all????
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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YT2095
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ok no Takers, here`s what I have.
I mole of KClO3 = 122.5 Grams
at STP the factor for 1 Litre (dm^3) is 0.0446
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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Xenoid
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OK!.. I'll have a go!
The gram molecular weight of oxygen occupies 22.4 litres at STP.
245.2g KClO3 ---> 149.2gKCl + 96gO2
1 litre oxygen at STP = 32/22.4 = 1.429g
Thus weight KClO3 required = 1.429/96 *245.2
Which equals 3.65g KClO3 to give 1 litre O2
Doesn't seem like much does it!
Regards, Xenoid
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woelen
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This computation can be done even more easily.
1 liter of (ideal) gas at STP is 1/22.4 mol of molecules is 0.0446 mol of molecules.
2KClO3 --> 2KCl + 3O2
So, you need (2/3)*0.0446 = 0.0298 mol of KClO3, which equals 3.65 grams.
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YT2095
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WOW, I`m really impressed! seriously I am, but at the same time I can`t understand Why it makes no sense to me in terms of picture`s/realist?
I don`t know when to use the divide or multiply???
people seem to use X when I would use /
I need it to make sense to me in order for me to remember it
if I can`t "see" it, I can`t remember it
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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woelen
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Try to explain more explicitly, where you have difficulty deciding when to use division or multiplication. This a rather basic math, and you should be
fluent in it. It is a matter of simply doing lots of this, although that may be very boring. Experience makes things easier.
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The_Davster
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I do it the same way as woelen, but I will format differently in the way I learned
2KClO3................---> 2KCl...............+ 3O2
.........................................................V=1L
..........................................................n=1/22.4
........................................................ n=0.0446mol
n=n(of O2)*(ratio of chlorate to O2)
n=(0.0446)*(2/3)
n=0.0298 mol
n=m/M
m=n*M
m=3.65g
[Edited on 13-11-2007 by The_Davster]
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Xenoid
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As I always tell my kids;
If something equals something else then whatever you do to one side of the equation you must do to the other to keep them equal.
For example if something equals 5.67g and you want to find out what equals 1g you have to divide the 5.67 by 5.67 to get 1, therfore you also have to
divide the "something" on the other side of the equation by 5.67 as well.
It's important to understand cross multiplying in calculations like these, for example if;
X / 35.78 = 45.67 / 32.89
then X = 45.67 / 32.89 * 35.78
because to change X / 35.78 to X you have to multiply by 35.78
Thus endeth Math 101, note, the numbers above are random and don't refer to the previous calculations.
Regards, Xenoid
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Magpie
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I will add my attempt to help:
| Quote: |
2KClO3 ---> 2KCl + 3O2
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This is a problem in stoichiometry and as already stated by woelen facility in solving such problems is very important. I face a similar problem
everytime I set up a chemistry experiment in my home lab, unless it is completely from a cookbook.
It is also a problem in proportions and I think that this is the clearest way to go about it:
1. You start off knowing 1 mole (32g) of diatomic oxygen has a volume of 22.4 liters at STP.
2. From the equation above, by proportions, you can see that for every mole of oxygen you will need 2/3's mole of KClO3.
3. You need 1 liter of oxygen therefore you need 1/22.4 moles of oxygen.
4. For the KClO3 you need (2/3)(1/22.4) moles.
5. Grams KClO3 then = (MW KCLO3 in grams)(2/3)(1/22.4).
This may not be the simplest way but I'm hoping it will help you visualize what is happening and why.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Bolt
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Um... maybe I haven't properly understood YT's question, but I see that when he asks about whether to multiply or divide, he's asking about
conversions.
Sometimes I get more advanced conversions confused myself.
You need to look at the UNITS of your scalar quanitites.
Thus, if you have 10 g/mol and you want an end result in grams, you need to multiply by moles. That way the moles will cancel out and you will be left
with grams.
The same thing applies with mol/L, g/L, etc.
For long conversions, use this format. (E.g, in the conversion of Watts to kJ/s.)
Starting units: 250,000 W
250,000 W |.........1 kW.........|.......1 kJ/s .....|
---------------|---------------------|-----------------| = 250 kJ/s
........1..........|......1000 W.......|........1 kW.......|
Disregard the periods.
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WizardX
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PV = nRT
n = RT/PV
Future calculations use this...
Standard Temperature and Pressure Calculator http://www.1728.com/stp.htm
STP or standard temperature and pressure http://en.wikipedia.org/wiki/Standard_temperature_and_pressu...
2 KClO3 ==> 2 KCl + 3 O2
Since you need 1 mole of O2 = 1/3, because 3/3 = 1
Therefore, 1/3 x 2 = 0.67 moles of KClO3
Moles to Grams: M x n = m therefore 122.55 x 0.67 = 82.1085 grams of KClO3
Albert Einstein - \"Great ideas often receive violent opposition from mediocre minds.\"
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Xenoid
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| Quote: | Originally posted by WizardX
Since you need 1 mole of O2 = 1/3, because 3/3 = 1
Therefore, 1/3 x 2 = 0.67 moles of KClO3
Moles to Grams: M x n = m therefore 122.55 x 0.67 = 82.1085 grams of KClO3 |
Yes, but YT2095 doesn't want 1 mole of O2 ........ (22.4 litres)
He only wants 1 litre
Therefore 82.11g / 22.4 = 3.66g (almost right... )
Regards, Xenoid
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WizardX
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Oversight!
| Quote: | Originally posted by Xenoid
| Quote: | Originally posted by WizardX
Since you need 1 mole of O2 = 1/3, because 3/3 = 1
Therefore, 1/3 x 2 = 0.67 moles of KClO3
Moles to Grams: M x n = m therefore 122.55 x 0.67 = 82.1085 grams of KClO3 |
Yes, but YT2095 doesn't want 1 mole of O2 ........ (22.4 litres)
He only wants 1 litre
Therefore 82.11g / 22.4 = 3.66g (almost right... )
Regards, Xenoid |
Thanks for correcting this oversight! Multi-tasking = errors.
Moles of O2 at STP = 0.0446
KClO3:O2 = 2/3
2:3
2/3 x 0.0446 = 0.03 moles of KClO3
0.03 x 122.55 = 3.6765 grams of KClO3
Albert Einstein - \"Great ideas often receive violent opposition from mediocre minds.\"
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YT2095
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well One of the problems I got in trouble with was thinking that 3 lots of O2 is the same as 2 lots of O3 as in 2KClO3.
so I didn`t use the 2:3 ratio, I used 1:1 thinking they were both the same.
the other is when I want say 2 litres at STP, I would Think 22.4 divide by 2.
as I said it`s a mental block thing I seem to have and can`t figure a way past it.
and don`t even get me started on Percentages
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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Ramiel
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One thing that they 'taught' at eng. 101 was conservation (or was it conversion!) of units, and it always seemed like they could have taught us kids
that earlier (like primary school!). Anyway, if you have something in L, and want it in g, then you multiply by g/L
g = L . (g / L)
seems simple enough! 
this came literally without thinking for me, but I'm aware that most people in my class had to labor over it to 'get it'.
hopefully this will help you get it.
Edit - one thing that really helps in cementing this principle is to list the units you're using AFTER EVERY STEP, even in working. This way you can
'see' where you are in a calculation... and where you need to get to. Many a time I've done physics questions by just applying laws to a unit until I
get the needed units (usually without knowing the material   )
[Edited on 14-11-2007 by Ramiel]
Caveat Orator
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YT2095
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a way I can see that Does make some kind of sense to me is this, if we just take the Question and forget the formula and stuff.
we know that 1 mole of KClO3 when heated will give 1.5 moles of O2.
so 22.4 x 1.5 = 33.6 litres or oxygen from a mole of the chlorate.
and 1 mole of KClO3 is 122.5g
so 122.5 / 33.6 = 3.645 g
this way there`s no need to complicate stuff with ratios and 0.0446 and that other stuff, and there`s no RXN equation needed either.
I`ve sat here and read and re-read these posts, and am more convinced than ever I`m a complete dunce at maths, so I tried to think deeply about what
the question wants and boil it down into something Simple for me to understand, and I came up with the above.
can someone check it for me to make sure it`s Right please, I think my answer is a little different after a few decimal places than the others.
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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YT2095
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I`m going to try one with Perchlorate and see if it works again, we`ll assume the question is the same and I want 1 litre of O2 and need to know how
many grams of of KClO4 I need.
1 mole of KClO4 will make me 2 moles of O2, and that is 22.4 x 2 = 44.8 litres of gas.
a mole of KClO4 weighs 138.5 grams
so 138.5 / 44.8 = 3.09 grams of it to make my 1 litre of O2 gas.
is that right?
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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woelen
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Yes, this also is one way to solve this kind of problems. And indeed, many questions can be answered in multiple ways, leading to the same final
answers. Which way is the best? That one, which suits you and works for you!
=================================================================
Now, let's add a somewhat more demanding thing, but still not that difficult. You should be able to solve that one. If you can do this, then you
certainly will pass exams with this kind of questions.
If you mix thionyl chloride with water, you get the following:
SOCl2 + H2O --> SO2 + 2HCl (two liquids are mixed and only gases are produced).
Assuming that both produced gases are ideal, then how many milliliters of water and how many milliliters of thionyl chloride are needed to produce 1
liter of gas at standard temperature and pressure?
In order to solve this one, you need to do a small Internet search. I'll provide you with the link, which contains the information you need in order
to solve the above question, so that you can concentrate on the math and not on the google search:
http://en.wikipedia.org/wiki/Thionyl_chloride
Which information you need, that is something which you should decide.
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YT2095
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I work out that you need a total of 9.168g of total reactants.
the SO2 side I get 6.11 and the HCl side I get 3.058
do those numbers make any sense at all?????
edit: here`s my working out;
SOCl2 = 119g/mol
H2O = 18g/mol
the total is 137 g/mol
it makes 1 mol of SO2 and that 22.4 Litres
and makes 2 mols of HCl and that`s 44.8 litres
so 137 / 44.8 = 3.058
and 137 / 22.4 = 6.11
that`s as far as I got.
I think I know what I`m trying to do, but I lose myself with maths and just freeze, then I forget my plan and have to start again
[Edited on 14-11-2007 by YT2095]
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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woelen
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No, these numbers are not correct.
SOCl2 + H2O --> SO2 + 2HCl
So, 1 mole of SOCl2 and 1 mole of H2O produce three moles of gas mix (remember, non-reacting gases may be added up). So, for making 1 mole of gas mix,
you need 1/3 mole of SOCl2 and 1/3 mole of water.
You want 1/22.4 mole of gas-mix (1 liter), so you need to divide by 22.4, and hence you need
1/3/22.4 mole of SOCl2 and
1/3/22.4 mole of H2O
Total grams of SOCl2: 119/3/22.4 = 1.77 grams
Total grams of H2O: 18/3/22.4 = 0.268 grams
Now you need to convert to ml. The info, needed for that is in the Wiki page I gave. I assume that you can complete that part.
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YT2095
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1.77g
and liquid is 1.638 g/ml
so I need just over 1 ml to make a litre.
I have No idea how to make them fit
it`s just over 1ml to me, I need 0.132 extra grams to make it work.
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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woelen
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If one ml is 1.638 gram, then we have that 1 gram is 1/1.638 ml.
Now it is easy to compute the volume of 1.77 gram. This is 1.77 times as much as the volume of 1 gram, hence that volume is 1.77*(1/1.638) =
1.77/1.638 = 1.081 ml.
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