[mikasaur]

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 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.

[AWK]

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.

[jamesbrown]

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

[sjb]

Does it make sense that the mass you end up with is greater than that you start with?

[jamesbrown]

Thats what I was thinking but where have i gone wrong

[AWK]

mass of 2Al2O3 = 203.922

[jamesbrown]

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?

[AWK]

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.

[mikasaur]

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 M_p is the mass of the product
M_r is the mass of the reactant
X is the stoichiometric coefficient of the product
MW_p is the molecular weight of the product
Y is the stoichiometric coefficient of the reactant
MW_r 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).

[aga]

These may or may not be useful.