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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: jamesbrown on January 23, 2016, 12:14:13 PM

Title: Calculating Amounts of substances Moles
Post by: jamesbrown on January 23, 2016, 12:14:13 PM
Hello I was just wanting to check if I got this question right.

How much sulphur dioxide would be made from 320 tonnes of sulfur.
If the equation is S+O2=SO2
that would mean the ratio would be 1 : 1 : 2
Also the mass would be 32 : 32 : 64
Which means that i would have to double the amount of Sulphur so it would be 320*2 = 640 tonnes
Is that right?
Title: Re: Calculating Amounts of substances Moles
Post by: Hunter2 on January 23, 2016, 01:25:19 PM
Yes corrrect. Better is to calculate via mole. 320 t Sulphur  corresponds 10000 kmol.  This correspond also to the SO2 . With the molar mass of 64 kg/kmol gives 640 t.
Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 23, 2016, 01:49:38 PM
If 48 tonnes of iron ore is used in a blast furnace what would the the mass of iron be. If the equation is:
Fe2O3 + 3CO = 2Fe + 3CO2

Would I find the moles of iron oxide.
But I'm not really sure what to do after that?

Can someone help please It would be really helpful.

Thanks
Title: Re: Calculating Amounts of substances Moles
Post by: Hunter2 on January 23, 2016, 02:17:33 PM
You calculate the moles of the 48 t Iron oxide.

If you have that then you can see the ratio of the moles of iron oxide to iron in the equation.

With the new mole value you calculate backwards to the mass of iron.

You need the molar masses of iron oxide and iron itself.
Title: Re: Calculating Amounts of substances Moles
Post by: AWK on January 23, 2016, 04:19:52 PM
Just read the balanced reaction - one mole of Fe2O3 gives 2 moles of Fe.
Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 23, 2016, 04:23:51 PM
So the moles of Iron oxide is 300000
So does that mean iron has 120000 moles
and oxygen has 180000 moles
and because in the equation we are making 2 iron does that mean the moles of iron is 120000*2=240000
then we need to find the mass of it so would it be i don't know could someone help?
Title: Re: Calculating Amounts of substances Moles
Post by: Hunter2 on January 23, 2016, 04:42:40 PM
No you get 300000 mole Fe2O3. This correspond according the equation to the double amount of iron. So the moles of Fe is 600000 moles.

This you have to calculate back to the mass of iron.
Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 23, 2016, 05:56:27 PM
Can someone just give me a step by step method on how do do mole questions like these. I'm really trying to understand but Its quite complicated in my opinion. Here are a list of the question that I have to do. So yeah. Thank you for trying to help someone dumb like me.(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2F6uL8WVh.jpg&hash=1416232c77d56cff469227ded1df1a9d285b544a)
Title: Re: Calculating Amounts of substances Moles
Post by: Burner on January 23, 2016, 10:02:55 PM
If I am not mistaken,

(Number of moles)=(mass)/(molar mass)

So, (mass)=(Number of moles)*(molar mass)

Moreover, we uses mole ratio directly from the balanced equation. Take question 2 as an example, you have 300000 mol of Fe2O3. From the balanced equation, mole ratio of Fe2O3 and Fe is 1:2, i.e. 1 mol of Fe2O3 gives 2 mol of Fe. Thus, 300000 mol of Fe2O3 gives 600000 mol of Fe. Then, you have to multiply that 600000 mol of Fe with its molar mass, which is 56. Finally, you get 600000*56=33600000g of Fe, or 33.6 tonnes of Fe.

Please kindly correct me if I have made any mistakes.
Title: Re: Calculating Amounts of substances Moles
Post by: Hunter2 on January 24, 2016, 07:01:56 AM
Generally the steps are:
1. Write chemical equation if it is not given.
2. Calculate the given mass into moles as described n = m/M, n [mole], m [mass] and M [molar mass]
3. Check how much moles correspond to the material what you looking for (Fe2O3 to Fe here 1:2)
4. Calculate back to the mass with the formula given in 2.
Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 24, 2016, 08:13:20 AM
For question 3 I have worked out the amount of moles for sodium chloride which is 136752.1368
but then for the ratio would it be 3:2:1 so i would have 91168.0912 moles of sodium and then i would have 45584.0456 mole of chlorine.
Is this right I don't know.
Then if this is right what would i do next?
Title: Re: Calculating Amounts of substances Moles
Post by: Hunter2 on January 24, 2016, 08:29:25 AM
First calculate the mole NaCl  n = 16000 kg/58,5 kg/Kmol =273,5 kmol

In the equation you can see that the mole amount of sodium is the same like the NaCl. So the given moles multiplited with the molar mass of sodium gives 6290,59 kg

For the Chlorine you see its the half it gives then 9695,73 kg
Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 24, 2016, 08:53:50 AM
Hunter why did you not convert the mas into grams and why did you not work out the moles for 2nacl because you only worked out the moles for nacl
Title: Re: Calculating Amounts of substances Moles
Post by: Hunter2 on January 24, 2016, 09:00:34 AM
What do you mean, I didn't convert the mass in gram. The mass is given in gram. 16 t = 16000 kg and this is equal to 273,5 kmol.

The equation is 2 NaCl => 2 Na + Cl2

So 2 NaCl correspond to 2 Na. This  is the same like NaCl correspond to Na. You can erase the 2 on both sides.

And 2 NaCl to Cl2 is the same like NaCl to 0.5 Cl2
Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 24, 2016, 09:04:18 AM
Sorry i meant why did you convert it to kg not grams
Title: Re: Calculating Amounts of substances Moles
Post by: Hunter2 on January 24, 2016, 09:06:37 AM
I do not like to write so long numbers.

1 kilogram = 1000 g. The same with moles 1 kmol = 1000 mol. But it doesnt matter at the end you had to get back to kilogram or tons.
Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 24, 2016, 10:19:49 AM
I am so sorry but I really don't understand any of the things you have been talking about could you please explain how you got 9694,73 kg of chlorine and do it in grams please. Also could you explain how I would do the next step which is ratio's i'm guessing? Also Thanks again for putting up with my stupidity as I know for you this is super easy but for me it is quite hard and complex.
Title: Re: Calculating Amounts of substances Moles
Post by: Hunter2 on January 24, 2016, 10:45:53 AM
We have the equation

2 NaCl => 2 Na + Cl2

what means 2 mole of sodiumchloride will convert to 2 mole of sodium and 1 mole of chlorine gas, which is a two-atomic molecule.

We have 16 t NaCl this must fit to x tons sodium + y tons chlorine.

Molar mass of NaCl is 58,5 g/mol, Na 23 g/mol and Cl2 71 g/mol  Units can be extended to kg/kmol
 
The moles must be equal

So u can write n = m(NaCl)/xM(NaCl) = m(Na)/yM(Na) = m(Cl2)/zM(Cl2)

x,y,z= stoechiometric coefficients. x = 2 , y = 2, z =1

m(NaCl) = 16 t = 16000 kg

n = 16000 kg/x 58,6 kg/kmol = 273,5 kmol/x

m (Na) = n *y M(Na)   m(Na) = 273, 5kmol/x * y* 23 kg/kmol = (y/x) 6290,5 kg, y =x = 2, Result 6290,5 kg

m(Cl2) = n*z(M(Cl2)  m(Cl2) =273,5 kmol/x * z * 71 kg/kmol =(z/x)*  19418,5 kg , x =2 z =1

m(Cl2) `=1/2 *19418,5 =9709,25 kg

6290,5 kg + 9709,25 kg =15999,75 kg ~ 16 t
Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 24, 2016, 11:41:14 AM
So the mass of Chlorien is 9709,25kg and the mass of Sodium is 6290,5kg
Thank you it is very complex to follow though so I am not certain about this yet.
Title: Re: Calculating Amounts of substances Moles
Post by: Hunter2 on January 24, 2016, 12:45:12 PM
To make it more easier

you can balance

m1/xM1 = m2/yM2

m1,2 mass

M1,2 molar mass

x,y stoechoimetric coeficients

Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 24, 2016, 01:26:36 PM
I get the mass and molar mass part and the equation but not quite sure what the x,y stoechoimetric coeficients are to be honest. Is it the ratio or what and could you tell me what it is for question 4 (so x,y)
Title: Re: Calculating Amounts of substances Moles
Post by: Hunter2 on January 24, 2016, 02:03:01 PM
You have C6H12O6 => 2 C2H5OH + 2 CO2

Molar Mass sugar is 180 g/mol , Ethanol 46 g/mol carbondioxide 44 g/mol

Coefficients are 1 for sugar and 2 for carbondioxide and alcohol.

2 kg/1*180 kg/kmol = xkg/2*46 kg/kmol

x = 1,022 kg Alcohole

2 kg /1*180 kg/kmol = ykg/2*44 kg/kmol

y= 0,977 kg

1,022 kg + 0,977 kg = 2kg
Title: Re: Calculating Amounts of substances Moles
Post by: mikasaur on January 25, 2016, 12:44:11 PM
I get the mass and molar mass part and the equation but not quite sure what the x,y stoechoimetric coeficients are to be honest. Is it the ratio or what and could you tell me what it is for question 4 (so x,y)

For questions like these, it's all about tracking your units. I find that writing out expressions where the units cancel is paramount to getting the question right.

C6H12O6  :rarrow: 2 C2H5OH + 2 CO2

MWsugar = 180 g/mol

MWalcohol = 46 g/mol

[itex]\textrm{2 kg sugar } \times\frac{1000\textrm{ g}}{1\textrm{ kg}}\times\frac{1\textrm{ mol sugar}}{180 \textrm{ g sugar}} \times\frac{2 \textrm{ mol alcohol}}{1 \textrm{ mol sugar}}\times \frac{46 \textrm{ g alcohol}}{1 \textrm{ mol alcohol}}[/itex]

See how in the above expression you can start to cancel out units left to right? If you want to always use grams and moles (rather than kg and kmoles) the first thing you do is convert from kg to g. Then you can convert from grams sugar to moles sugar. Then the (sorta) tricky part of using the stoichiometric coefficients. For every 1 mol of sugar consumed you create 2 moles of alcohol! Then you can convert the moles of alcohol to grams of alcohol, et voilà. You should be left with only the units you care about, which in this case is a mass of alcohol (in grams). I've left out the conversion back to kg if that's the unit you want.

I would set up your problems like this. Include the balanced chemical equation like Hunter2 said. Cancel units. Double check that the units that remain make sense.
Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 25, 2016, 01:07:22 PM
So for question 5 would you do this

The equation is 2Al2o3 => 4Al + 3O2

The molar mass for Aluminium Oxide is 204

4*27 = 108

1:0.52

5*0.52 = 2.6

Is this right or not?

Also thanks for helping me and I have no clue at all how to do question 6 if you could please help me do that.
Title: Re: Calculating Amounts of substances Moles
Post by: mikasaur on January 25, 2016, 01:14:59 PM
I don't know if that's right or not. It's hard for me to tell because you have no units in any of your numbers. The molar mass is 204 what? Why are you multiplying 4 by 27? What is the meaning of the number 108? Where does 0.52 come from?

Use units and set up an expression like I did. Cancel the units. Check your answer.

Hunter2 has already mentioned it, but the general approach for stoichiometry questions like these (where they give you a mass of reactant and the reaction goes to completion) is:

1. Write the balanced chemical equation
2. Convert the mass of reactant to moles of reactant
3. Use the stoichiometric coefficients to relate the moles of reactant to the moles of product
4. Convert the moles of product to mass of product
Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 25, 2016, 01:17:59 PM
Sorry man I have a list of questions on page 1 of the post. There you can do it yourself and tell me If I got it correct. Also I have no clue what to do with question 6. But anyway thanks for helping me.
Title: Re: Calculating Amounts of substances Moles
Post by: AWK on January 25, 2016, 01:31:40 PM
Molar mass of Al2O3 is 102
54/102=0.529
Final result in tonnes is an accepted value
Title: Re: Calculating Amounts of substances Moles
Post by: jamesbrown on January 25, 2016, 01:35:18 PM
So AWK is 2.645 tonnes the right answer
Title: Re: Calculating Amounts of substances Moles
Post by: Borek on January 25, 2016, 01:38:28 PM
Sorry man I have a list of questions on page 1 of the post. There you can do it yourself and tell me If I got it correct. Also I have no clue what to do with question 6. But anyway thanks for helping me.

Sorry if I will sound a bit rash, but if you want to learn try to listen to what others tell you. mikasaur is perfectly correct when he points out your numbers look quite random without units and are hard to follow - which is not OUR problem. It is YOUR problem because that's what makes you lost.

In most cases we have enough experience to guess what you are doing, so we can try to help - but what makes our work just harder makes your work impossible.
Title: Re: Calculating Amounts of substances Moles
Post by: mikasaur on January 25, 2016, 01:39:34 PM
Sorry man I have a list of questions on page 1 of the post. There you can do it yourself and tell me If I got it correct. Also I have no clue what to do with question 6. But anyway thanks for helping me.

I know you had a list of questions. When I ask for the units I'm asking if you know them. I know you're getting 27 from the MW of Al and that you're getting the 4 from the stoichiometric coefficient. But what you need to ask yourself is why one of the first things you need to do is multiply 4 by 27 (I don't think it is).

Use the four steps I outlined below, include units, and question 5 should be pretty simple. Question six is not terribly well-worded. I would assume that they mean 2% by mass.

My step 1 is a little harder here and can be skipped by someone well-versed in these sorts of problems but if I were you I would at least try to think about it. Do you think you could make a balanced chemical equation for the process they're describing? Hint: you may have to represent coal as SX, where X is the "stuff" in coal that isn't sulfur.
Title: Re: Calculating Amounts of substances Moles
Post by: AWK on January 25, 2016, 02:04:19 PM
I prefer solution of this kind of stoichiometric problems through mass fraction that can be treated as unitless value.
mass fraction of Al in Al2O3 is 2Al/Al2O3=54/102
Hence mass of Al in 5 tonnes Al2O3 (5 treated as an exact number)  is 0.529×5 tonnes with final 2 significant digits.
Title: Finding the Theoretical mass of Sulphur
Post by: jamesbrown on January 26, 2016, 04:23:52 PM
We put in 50ml of sodium thiosulfate in and we put 5ml of hydrochloric acid in.

This is the equation:
HCl + sodium thiosulfate   sodium chloride + sulfur dioxide + sulfur + water.
2HCl(aq)  +  Na2S2O3(aq)           2NaCl(aq)    +      SO2(g)    +   S(s)  +  H2O(l)

I was given this as an extension and have no idea how to do this as we haven't gone over anything to do with theoretical mass. So i'm really stuck, so if someone could work it out for me and explain it I would be hugely appreciative of it.

Thanks 
Title: Re: Finding the Theoretical mass of Sulphur
Post by: mikasaur on January 26, 2016, 04:25:46 PM
Is that all there is to the prompt? Was this a lab or part of a problem set?

If you give us the exact wording of the question we can help out a little more.
Title: Re: Finding the Theoretical mass of Sulphur
Post by: jamesbrown on January 26, 2016, 04:29:59 PM
We performed the test in a lab and here are a few graph that might be helpful. (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2FjLmmqwj.png&hash=6d5b01eadfd1aa6d477e6faf409432324f2c16ee) (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2FsdSQoYR.png&hash=747ad470e8ece8de2a7b36e0113bf83926cc1d44)
Title: Re: Finding the Theoretical mass of Sulphur
Post by: mikasaur on January 26, 2016, 04:36:04 PM
It's still very hard to help you with the information given.

I'm not sure what you mean by an "extension". Was this a question that was done after your lab? What exactly was the lab?
Title: Re: Finding the Theoretical mass of Sulphur
Post by: jamesbrown on January 26, 2016, 04:53:02 PM
Experiment:
We tested how quickly 50ml of Sodium thiosulphate solution reacted with 5ml of hydrochloric acid at different temperatures to test how temperatures affect rate of reaction.
Method:
Prepare 50 ml of thiosulphate and 5ml of HCL, stopwatch and paper with a clear X on it(and a conical flash to mix solutions in)
Then heat water in a beaker until the required temperatures, when this is done put the conical flask inside the beaker (using it as a water bath) place it on the paper then mix both solutions.
Then time how long it takes for the thiosulphate and HCl take to react by waiting until the cross on the paper is unreadable through the solution. (HCL and Thiosulphate make an opaque product)
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2FoQqGB8c.png&hash=66e64595469816d80edec04eddf0bb1e625e2af3)
Title: Re: Finding the Theoretical mass of Sulphur
Post by: Borek on January 26, 2016, 05:10:32 PM
Still not clear what is the "extension" you talk about.

Prepare 50 ml of thiosulphate and 5ml of HCL

You were not told anything about concentrations?
Title: Re: Finding the Theoretical mass of Sulphur
Post by: jamesbrown on January 26, 2016, 05:17:23 PM
Extension means an extra question

concentrations:
Sodium thiosulphate 0.15
Hydrochloric acid 2
Title: Re: Finding the Theoretical mass of Sulphur
Post by: mikasaur on January 26, 2016, 05:22:17 PM

concentrations:
Sodium thiosulphate 0.15
Hydrochloric acid 2

Units, dude!! Numbers without units are meaningless.
Title: Re: Finding the Theoretical mass of Sulphur
Post by: jamesbrown on January 26, 2016, 05:38:55 PM

concentrations:
Sodium thiosulphate 0.15
Hydrochloric acid 2

Units, dude!! Numbers without units are meaningless.

moles/dm3
Title: Re: Finding the Theoretical mass of Sulphur
Post by: mikasaur on January 26, 2016, 06:07:04 PM
Cool. There ya go.

So there was no prompt for your extension? You just got a piece of paper that said "find the theoretical mass of sulphur"? I feel like we're missing something here.
Title: Re: Finding the Theoretical mass of Sulphur
Post by: AWK on January 26, 2016, 06:22:02 PM
Seems to be a stoichiometric problem with a limiting reagent.
Title: Re: Finding the Theoretical mass of Sulphur
Post by: Burner on January 27, 2016, 12:33:28 AM
So there was no prompt for your extension? You just got a piece of paper that said "find the theoretical mass of sulphur"? I feel like we're missing something here.

I think the paper maybe asking you to calculate the percentage yield based on the theoretical mass and the actual mass of sulphur obtained, as that's what our pubilc exam papers(HKDSE) always do.

Anyway, I think that calculating theoretical mass is simliar to ordinary mole calculations involving mass, though in this lab we may need to find out the limiting reagent in the reaction and use that number of moles of that limiting reagent to do calculations.
Title: Re: Finding the Theoretical mass of Sulphur
Post by: Borek on January 27, 2016, 03:16:07 AM
I suspected the type of problem AWK named, which is why I asked if you are given concentrations. If so, the question to solve is most likely just "What is the theoretical mass of sulfur produced when you mix 50 mL of 0.15 M sodium thiosulfate with 5 mL of 2 M HCl solution".
Title: Re: Finding the Theoretical mass of Sulphur
Post by: jamesbrown on January 27, 2016, 01:23:34 PM
So do you guys know how to do it or not?

Thank you for this by the way.
Title: Re: Finding the Theoretical mass of Sulphur
Post by: mikasaur on January 27, 2016, 01:36:38 PM
So do you guys know how to do it or not?

Thank you for this by the way.

We still don't know the exact question your teacher is asking. AWK, Burner, and Borek are all very intelligent and experienced so they've figured out what the question probably is. We're all trying to help, but I would advise trying to learn how to ask questions well. This sort of question shouldn't need a dozen responses to answer. Not trying to harsh, but that's the reality of it.

Anyways if the question really is "What is the theoretical mass of sulfur produced when you mix 50 mL of 0.15 M sodium thiosulfate with 5 mL of 2 M HCl solution" then it's a stoichiometry problem with a limiting reagent.

Have you learned about limiting reagents? They're very similar to the questions you were asking about here (http://www.chemicalforums.com/index.php?topic=84071.0) except you need to find out which of your two reagents are completely used up and which one isn't. From there it's basically the same.

To get you started, if you had a million moles of sodium thiosulfate and 0.0001 moles of HCl, which do you think would limit the reaction? Now use that same concept but do so quantitatively with your actual values.
Title: Re: Finding the Theoretical mass of Sulphur
Post by: AWK on January 27, 2016, 01:39:51 PM
Just solve the stoichiometric problem with a limiting reagent. You have a balanced reaction. You have to check which reagent is in excess - HCl or Na2S2O3.
Title: Check my Mole Equations
Post by: jamesbrown on January 31, 2016, 10:56:01 AM
Hi I have been practicing my moles questions and I was just wondering if they were correct or not.

2Al2O3 -> 4Al + 3O2
Calculate the mass of aluminium that would be made from 5 tonnes of aluminium oxide.

moles = mass/mr
moles = 5000000g/156 (156 is the mr of 2Al2O3)
moles = 32051.2821

Al2O3 : Al
1mole:2mole
32051.2821:64102.5641

mass = moles * mr
mass = 64102.5641 * 108(mr of 4Al)
mass = 6923076.92g

mass = 6.92 tonnes

Is this right? Thank you
Title: Re: Check my Mole Equations
Post by: sjb on January 31, 2016, 11:22:51 AM
Does it make sense that the mass you end up with is greater than that you start with?
Title: Re: Check my Mole Equations
Post by: jamesbrown on January 31, 2016, 11:48:23 AM
Thats what I was thinking but where have i gone wrong
Title: Re: Check my Mole Equations
Post by: AWK on January 31, 2016, 12:30:57 PM
mass of 2Al2O3 = 203.922
Title: Re: Check my Mole Equations
Post by: jamesbrown on January 31, 2016, 03:59:47 PM
But i thought 2Al2O3 was

Al  = 27
O = 16
O3 = 48
Al2 = 54

I thought the 2 on the 2Al2 was only for the al2 not the O3 as well so:

48+54 = 102
102*2 = 204

moles = mass/mr
moles = 5000000g/204 (204 is the mr of 2Al2O3)
moles = 24509.8039

Al2O3 : Al
1mole:2mole
24509.8039:49019.6078

mass = moles * mr
mass = 49019.6078 * 108(mr of 4Al)
mass = 5294117.64g

mass = 5.29 tonnes


Thank you for this
Also do you think this is right?
Title: Re: Calculating Amounts of substances Moles
Post by: AWK on January 31, 2016, 06:07:38 PM
Still wrong - from 5 tonns of oxide you cannot obtain, eg 6 tonns of pure metals.

I think you may calculate percentage of metal in your oxide, then in bulk sample.
This also gives you correct amount of metal.
Title: Re: Calculating Amounts of substances Moles
Post by: mikasaur on January 31, 2016, 07:07:09 PM
I think your problem is that you're kind of mixing two different approaches to solving stoichiometric problems.

The shortcut way for simple problems like this is to use mass fractions (I'm not sure if that's what it's actually called). In this example:

[itex]M_p = M_r \times \frac{X\cdot MW_p}{Y \cdot MW_r} [/itex]

Where Mp is the mass of the product
Mr is the mass of the reactant
X is the stoichiometric coefficient of the product
MWp is the molecular weight of the product
Y is the stoichiometric coefficient of the reactant
MWr is the molecular weight of the reactant

This method is quick and good for people who are well-practised in solving stoichiometry problems. The way that I would do it is slower, more methodical, but arguably less prone to error. I've said it before but you seem to have the basics:

1. Find the number of moles of reactant you have (use the molar mass of the actual reactant, ignoring coefficients)
2. Find the number of moles of product you'll produce using the stoichiometric coefficients (in this case, every 2 moles of oxide consumed creates 4 moles of pure metal)
3. Convert moles of product to mass of product

Your problem comes from including the 2 in the molecular mass of Al2O3 to get a molecular mass of 204 and THEN you multiply by 2 AGAIN in your stoichiometric relationship step (when you get 24.5k -> 49k moles). You're double counting and getting an answer that is too big (by two).
Title: Re: Calculating Amounts of substances Moles
Post by: aga on February 07, 2016, 04:38:15 PM
These may or may not be useful.