12.8 gr of O Flowed over 100 gr hot Al.
and I got this reactions equation: 4Al(s)+3O2(g)--->2Al2O3(g)

a)how many gr Aluminum oxide Al203 Created through a chemical process ?

b)how many unit of Al Were obtained through a chemical process

c)Which responds to what quantity of material left at the end of the process
thanks for help :)

a)

1. Find the number of moles of oxygen which flowed.
2. Find the number of moles of Al present.
3. Deduce the limiting reactant by comparing the mole ratio in the equation reaction.
4. From there, work out the number of moles of Al2O3 produced.
5. Find the equivalent in grams.

b) & c) I'm not sure I understand the question here... =S

Not that it affects the answer, but the Al2O3 product will be a solid, not a gas. That is why aluminum makes such a great rocket fuel - you get back the heat of vaporization and heat of fusion. Hotter flame, more kinetic energy, MORE THRUST!

OK well I got this:
1)0.8 moles of oxygen which flowed
2) I don't understand u mean the number of moles of Al in the Deliverables side of the equation ?
3. Deduce the limiting reactant by comparing the mole ratio in the equation reaction can u give me any direction ?

b)how many unit of Aluminum oxide was Created in the process?

c)What is material reacts and What amount in gr are left from the same material in the end of the Process
hope now is more Clear.

Look at your equation, you say you have 0.8 moles of oxygen but what is the correct formula for the oxygen present in the air (hint: it's correct in your equation) You DO NOT have 0.8 moles of oxygen. Why don't you know how many; moles of Al you have? You have grams so you have moles. When the reagents combine, which one will be used up first? That's the limiiting reagent. The equation tells how many moles of each will be needed. That tells how many moles of Al2O3 will be produced. Moles will give you grams. It's pretty obvious you will run out of oxygen before you run out of aluminum; subtract the grams of aluminum actually used from the 100 grams you start with and you get the amount of unreacted aluminum left over.

well the correct formula for the oxygen present isO2 or O3 so the number of moles of oxygen which flowed is 12.8/32=0.375 moles or 12.8/48=0.26 moles. (I think is need to be for O2).
when u told present u mean the Right side of the equation? I can know the number of moles of Al but just let me know when u told present what u mean. In the left side where I have 4Al so I have Mm=108g/mol m=100 so the number of the moles will be n=100/108?

thanks.

Um... I think that you have got some concepts wrong...

First of all, in your equation, you are already told that the oxygen is O_2 and not O_3

'the oxygen present in the air' this only means the oxygen that you have in the air, the oxygen that you breathe.

In your equation, it is written 4Al. This means that 4 moles of Al will react. This doesn't mean that Al_4 will react... this doesn't even 'exist'. So, we get:
M_m = 27\ g/mol


b)how many unit of Aluminum oxide was Created in the process?

Um... I'm not sure what does 'unit' refer to exactly... It seems really strange since in the previous part, you already found the number of moles of Al2O3 and from this calculated the mass of Al2O3 produced...


c)What is material reacts and What amount in gr are left from the same material in the end of the Process
hope now is more Clear.

This is still not so clear...

From your calculations, you should see that oxygen reacts completely and aluminium remains/is left behind.

You only need to work out the number of moles of Al used, and subtract this from the initial number of moles of Al, then find the mass.

WELL the number of moles of oxygen which flowed is 0.4 I do like this 12.8g/32 =0.4 according to this equation n=m/Mm
the number of moles of Al present (here I not sure)100g/27=3.7 moles.
the mole ratio in the equation 4:3:2.
tell me if I Right till here ***.
THANKS.

Yes, that's good so far :)

how I find the number of the moles of Aluminum oxide Al203 that Created through a chemical process?
u told me to Deduce the limiting reactant by comparing the mole ratio in the equation reaction. But what mole ratio ? Ratio between O2 and Al or?


thanks.

Yes you compare between the mole ratio of Al and O2

You have 3.7 mol Al and 0.4 mol of O2.

From your equation, 4 moles of Al react with 3 moles of O2

So, 0.4 mol of O2 would react with (4/3) x 0.4 = 0.5333 moles of Al.

Since you have 3.7 moles of Al, you have Al in excess. Hence, the number of used moles is 0.4 mol of O2 and 0.533 mol of Al.

How many moles of Al2O3 would they yield from your equation?

I Understand most of what u told.

I have 3.17 moles of Al in excess.
the number of moles of Al2O3 from my equation is 0.26?
I did like this 0.4*2/3. yes?
or I can do 0.533382/4=0.26 moles of Al2O3 hope I did good :)

Yes, that's it, but the actual number is 0.26666666... which becomes 0.267 moles.

And I guess you did a typo on this line and forgot to add times 2:


or I can do 0.533382/4=0.26 moles of Al2O3

Yes, now find the mass of this amount of moles of Al2O3. :)

well: 0.4*2/3=0.2666 moles of Al2O3 and the mass is 101.93*0.2666=27.17g or if I round down this result I will get 27g Al2O3.

YES?

THANKS SO MUCH ABOUT UR HELP

Yes :)

Really thanks you.