How Many Moles of Aluminum Are Needed to Form 3.4 Moles of Aluminum Oxide?

[AngelShare]

Aluminum oxide is formed from the reaction of metallic aluminum with oxygen gas. How many moles of Aluminum are needed to form 3.4 moles of Aluminum oxide?

I don't necessarily need help solving the problem itself (I want to try to do it on my own) but I don't understand how to set the equation up. Aluminum oxide? --> AlO? Metallic aluminum? --> Al? Oxygen gas? --> O?

Al + O --> AlO ?

That seems too easy...and wrong...

 

[Borek]

Look at the periodic table - Na₂O, MgO, SiO₂ - what should be the formula of aliminum oxide?

 

[AngelShare]

Sorry for the wait, computer problems. :)

Anyways, is it...

Al + O2 --> Al2O3

 

[Borek]

Exactly.

Now it should be a breeze...

 

[AngelShare]

Well, we're talking about me doing this so we'll see...

Is this correctly balanced?

4Al + 3O2 --> 2Al2O3

 

[AngelShare]

Ah, I know it was supposed to be easy (And I'm sure it is as my teacher said so as well on the lesson page.) but I don't know what to do. I think the problem lies in what it is I'm doing exactly...what is stoichiometry? I read the definitions and took the notes but it's still a bit fuzzy.

 

[Borek]

Reaction is balanced. Now - think in terms of proportions. You need 4 moles of Al to obtain 2 moles of oxide. If so, how many moles of Al are needed to prepare 3.4 mole of oxide?

 

[AngelShare]

I can do that but I don't understand why I'm doing it if I need to know how many moles of Aluminum are needed to make 3.4 moles of Aluminum Oxide...

 

[Gokul43201]

I can do that but I don't understand why I'm doing it if I need to know how many moles of Aluminum are needed to make 3.4 moles of Aluminum Oxide...

Perhaps you haven't thought about what the balanced equation is telling you. How would you put the equation into words ?

 

[AngelShare]

I'd probably say, "Four atoms of Aluminum and six atoms of Oxygen make Aluminum Oxide"...which seems to be quite wrong.

 

[Gokul43201]

I'd probably say, "Four atoms of Aluminum and six atoms of Oxygen make Aluminum Oxide"...which seems to be quite wrong.

Try "4 atoms of Aluminum react with 3 molecules of Oxygen, making two molecules of Aluminum Oxide".

The lesson to learn from this (that's relevant to the question asked) is that 4 atoms of Al give rise to 2 molecules of Al2O3.

There ! Piece of cake, eh ?

 

[AngelShare]

So you aren't counting the two after the O? If not, why? I was which is how I got confused.

 

[AngelShare]

Determine the number of grams of NH3 produced by the reaction of 3.5g of hydrogen gas with sufficient nitrogen gas. (Check for diatomic elements).

I wrote down the diatomic elements but I'm not sure if this is right:

H2 + N2 --> NH3

 

[Math Is Hard]

So you aren't counting the two after the O? If not, why? I was which is how I got confused.

note that he said 3 molecules of Oxygen, not atoms.

 

[AngelShare]

note that he said 3 molecules of Oxygen, not atoms.

Ah, okay. And that's done with Oxygen and not Aluminum because Oxygen is a diatomic element?

 

[Math Is Hard]

Ah, okay. And that's done with Oxygen and not Aluminum because Oxygen is a diatomic element?

yes, indeedy. Don't forget those diatomic elements --
Nobody Has Fear Of Ice Cold Beer.
And you can see there on the left side of the equation that your Al doesn't have a subscript like the O₂.

 

[AngelShare]

So, when creating an equation like that one, the diatomic element involved (If there is one.) always has a subscript of two, right? The other element needs no such thing? So, if I'm right about that, my other equation is correct?

H2 + N2 --> NH3

Hydrogen and Nitrogen are both diatomic elements so they both require the subscripted two? However, I'm not sure why I ended up with what I did in the first equation...

Al + O2 --> Al2O3

The product switched subscripts...Al was three and Oxygen was two at the beginning. Why is that exactly? I remembered doing that with my tutor last year which is why I did it here but why I did it isn't clear to me.

 

[Math Is Hard]

So, when creating an equation like that one, the diatomic element involved (If there is one.) always has a subscript of two, right? The other element needs no such thing? So, if I'm right about that, my other equation is correct?

H2 + N2 --> NH3

yes, but you will need to balance it.

Hydrogen and Nitrogen are both diatomic elements so they both require the subscripted two? However, I'm not sure why I ended up with what I did in the first equation...

Al + O2 --> Al2O3

The product switched subscripts...Al was three and Oxygen was two at the beginning. Why is that exactly? I remembered doing that with my tutor last year which is why I did it here but why I did it isn't clear to me.

well, you started with this
Al + O₂ --> Al₂O₃
but then you used coefficients to balance it, so your subscripts did not change, just the coefficients.
4Al + 3O₂ --> 2Al₂O₃

 

[Math Is Hard]

maybe it was the step where you balanced the charges to get Al₂O₃ that is confusing you?

 

[AngelShare]

I was hoping I had gotten the second one correct (Balancing the equation usually isn't too hard for me but setting it up is a different story. ).

What confused me about the first one is that I started out with this:

Al + O2 --> Al2O3

On the left side of the equation, Al has no number near it. Oxygen has a two after it. The final product is Al2O3 which means Al now has a number after it. Where did it come from? I vaguely remember my tutor saying something about, "Switch the numbers..." which is kind of confusing being that it is a vague memory.

Now, I know I was given the product in this particular equation but, let's say I was given this:

Al + O2 --> ?

...and I had to write the final product. I'd have to get Al2O3, right? If so, how would I get that all on my own? Am I hitting on something I should know or am I unnecessarily confusing myself?

As for balancing the second equation, would it be:

3H2 + N2 --> 2NH3

How did you get the numbers subscripted in your other post?

 

[Math Is Hard]

To get the Al₂O₃ :
You know that you are going to have some aluminum ions and some oxygen ions bonding together for your product. But how many Al ions and how many O ions are going to hook up with each other? What you have to do is look at what the charge is on your Al ion and the charge on the O ion. The Al ion has a charge of +3 (the same as its group number) and the O ion has a charge of -2 (the group number minus 8). But you need to make the charges balance (cancel each other out). You can figure that the least common multiple of 2 and 3 is 6. If you have two Al ions, the overall charge is +6 and if you have three Oxygen ions the overall charge is -6. Now you know what the subscripts need to be:
Al₂O₃

It's much easier if you have something like Na and Cl, where the Na ion charge is +1 and the Cl ion charge is -1. It's just NaCl.

To do the subscripts, you can just put tags around your numbers for sub and /sub, each enclosed with []. If you hit quote to reply, you should be able to see what I did.

p.s. this might be helpful in understanding what's going on when you are writing the formulas for ionic compounds:
http://www.iun.edu/~cpanhd/C101webnotes/composition/formioncmpds.html

 

[Math Is Hard]

As for balancing the second equation, would it be:

3H2 + N2 --> 2NH3

yes, that's it.

 

[AngelShare]

1. 1.2 moles
2. 2.3 grams

Is that right?

 

[Math Is Hard]

1. 1.2 moles
Is that right?

No, that can't be correct. Remember that your balanced equation was this:
4Al + 3O₂ --> 2Al₂O₃
Think about what Borek said:

Reaction is balanced. Now - think in terms of proportions. You need 4 moles of Al to obtain 2 moles of aluminum oxide. If so, how many moles of Al are needed to prepare 3.4 mole of oxide?

 

[AngelShare]

This is what I was told: do not over think this one. Just use the mole ratio from the balanced equation. Grams do not enter into this one. It is as simple as it seems.

Okay, following the example on the lesson page (Step by step... )...

Al/Al₂O₃
4/2

He said that grams don't enter into this one so I simply divided 4 by 2...wait, no I didn't...what in the hell did I do? (My method was clearly a dumb one.)

4 to get 2

I need to get 3.4...

4/2 = 3.4/x
4x = 6.8
x = 1.7?

So, I got the second one right or is that one so far off you can't figure out how I even got that answer?

 

[AngelShare]

1. 6.8 moles

Aw God, number two is going to kill me. I tried following his steps...

Given quantity in grams--> change to moles----> multiply by ratio from balanced equation -----> convert wanted quantity from moles to grams.

But I have no idea what I'm doing. Seriously, how I missed that part of the lesson is beyond me. Somehow, I got it to 1.5 moles (There's a 99.9% chance that that's incorrect but, at this point, I really don't care anymore. ) and am attempting to complete the last step-- convert wanted quantity from moles to grams and I don't even know how to start.

My teacher said, "After that try to do the work to the best of your ability ..." so that's what I'm going to do. I'm going to throw something together before I get a zero because I'll get a least a few points this way. Could someone please show me the steps and explain this to me? If I fail this class again, I think I'll lose it.

2. 11.9 grams

How 'bout that?

 

[Math Is Hard]

aww..honey, you are going to make yourself crazy. 6.8 mol of aluminum looks good to me for that first one. Problems like that, to me anyway, are just like changing a recipe to make more or less of it. Do you cook? Do you ever double or half a recipe? If I wanted to whip up a batch of four moles of aluminum oxide, I would look at my basic recipe for 2 moles of aluminum oxide and think, hmm..I need to double everything, and I would multiply all the coefficients by two. But what we want in this case is 3.4 moles of aluminum oxide, so I can look for some number that I can multiply to the 2 coefficient of the aluminum oxide that would make it 3.4 and then multiply that to all the other coefficients in the equation.

But I am a better cook than chemist! I hope that makes sense.

Sorry, I have not had a chance to look at your second problem yet.

 

[AngelShare]

I know, I could feel my sanity slipping there.

My most common problem with Chemistry is that I'm usually missing something...just one or two things that would make everything click. Even with one on one tutoring, most things need to be explained to me at least two times for it to make sense.

I still don't quite get what it is I'm doing with all this. Hell, I don't even understand what a mole is completely. I read the definition so many times I got sick of it. I can find molar masses and molar ratios but what a mole is exactly makes no sense...

"The word 'mole' is used to denote a quantity of particles, 6.02 x 10^23 particles to be exact."

So, basically, does that mean that one atom of Hydrogen has 602,000,000,000,000,000,000,000 particles in it? And all of that weighs 1.0080 grams? My mind can't seem to grasp that because, if that's the same for every element (And I got it right...), that means that Carbon has 602,000,000,000,000,000,000,000 particles in it but weighs 12.01115 grams. Is my understanding of what a mole is off?

On to Stoichiometry...number one is asking me how much of one thing is needed to create another, right? Got it. It's supposed to be a simple question and, assuming I got it right, it wasn't too bad once I figured out what I was to be doing. (I stumbled across http://misterguch.brinkster.net/molecalculations.html" and used it. )

The second one is a bit tougher. It takes a lot to get to where I need to end up and the steps confuse me. I knew I had to convert grams to moles but I wasn't sure how to go about doing that. I thought I'd have to set up an equation so that grams would cancel out and leave me with moles but I couldn't figure out what I'd put in the equation...I basically got this:

3.5 grams * 1 mole

...and then got stuck. I thought about putting the weight of...whatever...underneath the 1 mole so that the grams would cancel out but then I thought, "What weight?" Should I use the weight of Aluminum, the weight of the product as a whole, or something else? Or am I entirely off? What I ended up doing is a mystery to me so I'll have to look back through my chicken scratch to figure it out and take proper notes...

I don't get why I'm converting this from grams to moles and then taking my final answer and converting it back to gra...ah, never mind, that part just slid into place. But now I don''t understand why this step is in there: "multiply by ratio from balanced equation".

I don't know how I'm going to remember all of this, really. Lessons 5.10 included several equations, several of which I had problems with. I didn't know when to use which equation...

Molarity = moles of solute/liters of solution

moles = grams/molar mass

MV = grams/molar mass

MV = grams/GMW

What in the hell? I had all of this at least two weeks ago, how does anyone remember this?

What's particularly hard about cyber schooling is that the lessons are planned before the school year starts so the teacher can't say, "Ah, well, the students are struggling to I'll alter the lesson plan." because they didn't even set it up. We started off covering two lessons a week, then it was upped to four a week so that we could reach the end of the lessons by the end of the year. A dumb idea if you ask me because I can't complete four lessons a week.

Just had to rant a bit there, off to bed I go...I've stressed myself out enough for today and I'm sick to boot so I need a bit of rest.:zzz:

 

[ksinclair13]

So, basically, does that mean that one atom of Hydrogen has 602,000,000,000,000,000,000,000 particles in it?

An atom of hydrogen can't contain particles (an atom is smaller than a particle). One mole of hydrogen contains 6.022 x 10²³ atoms. One mole of carbon contains 6.022 x 10²³ atoms. One mole of NO₂ contains 6.022 x 10²³ molecules of NO₂. One mole of pencils contains 6.022 x 10²³ pencils.

 

[zoobyshoe]

I still don't quite get what it is I'm doing with all this. Hell, I don't even understand what a mole is completely. I read the definition so many times I got sick of it. I can find molar masses and molar ratios but what a mole is exactly makes no sense...

"The word 'mole' is used to denote a quantity of particles, 6.02 x 10^23 particles to be exact."

So, basically, does that mean that one atom of Hydrogen has 602,000,000,000,000,000,000,000 particles in it? And all of that weighs 1.0080 grams? My mind can't seem to grasp that because, if that's the same for every element (And I got it right...), that means that Carbon has 602,000,000,000,000,000,000,000 particles in it but weighs 12.01115 grams. Is my understanding of what a mole is off?

A mole is an almost arbitrary specific number, much like the term: a dozen. The weight of one atom is too small to be useful in the lab. Instead it was decided it was much more usefull to weigh them in batches of 6.022 x 10²³. That's a massive number of atoms, but they're so tiny a mole of them only comes out to register on the scale of grams. A mole of carbon atoms weighs very nearly 12 grams. A mole of Oxygen atoms (the same number of them) weighs 16 grams.

By knowing the weight in grams of a given element, you also know to a very useful extent how many atoms of it you are working with. Check out the periodic table: the atomic weight of an element is always the weight of one mole of that element.

If you know two elements always combine at a certain ratio, like two hydrogens always combine with one oxygen, then you know because of knowing the weight of a mole of those elements, how many grams of each to put together to combine with no, or not much, extra ingredients left over. 2 x 1 mole of hydrogen (2x 1.008 = 2.016 grams of hydrogen) plus 1 mole of oxygen = 16 grams of oxygen.

That's what a mole is all about. You are always comparing the weights of that specific number of atoms (or molecules), so you can work with chemicals in terms of weight (grams, kilos etc). The weight of a mole of one thing is different than the weight of a mole of another, just like the weight of a dozen eggs is different than the weight of a dozen apples (going back to Math Is Hard's cooking analogy).

 

[Math Is Hard]

I took a crack at that second problem :

Determine the number of grams of NH₃ produced by the reaction of 3.5g of hydrogen gas with sufficient nitrogen gas. (Check for diatomic elements).

and came out with 19.7 g NH₃ as a solution.

Anyone want to check my math on this? It's been a while since I've done these.

 

[AngelShare]

You tried it too, Math Is Hard? I'm still working on it as well.

#2 is correct in the first part. In the second part use the molar ratio from the coefficients in the original equation and then use the molar mass (g) of NH3. The molar mass should be 2/3. Resubmit for a higher grade.

This is what I was told...and here is my work...

3.5 grams of H₂/1 * 1 mole of H₂/2.02 grams of H₂

3.5/2.02

1.7 moles

1.7(2/3) = 1.1 moles

1.1 * 3.5 grams = 3.85 grams

That's my newest attempt at the second problem. My other answer was 11.9 I believe. As far as I can tell, I messed up the molar ratio part...I think I was looking at the wrong problem.

 

[Math Is Hard]

how would you calculate the molar mass of 3H₂?
how would you calculate the molar mass of 2NH₃?

try that first.

 

[AngelShare]

3H₂ = 6.048
2NH₃ = 34.062

Is that right?

 

[Math Is Hard]

3H₂ = 6.048
2NH₃ = 34.062

Is that right?

Good. so your mass ratio of hydrogen to ammonia is
6.048g H₂ / 34.06g NH₃
or 0.1776 g H₂ / 1 g NH₃

so if I am thinking correctly, we know we have 0.1776 grams of hydrogen for 1 g of ammonia.
but we want to see the ratio for 3.5 g of hydrogen. now what?

 

[AngelShare]

now what?

That's a good question...

I don't really understand what you're doing there as it seems to be going a little away from simply using the equation/"road map" we were given.

Given quantity in grams--> change to moles----> multiply by ratio from balanced equation -----> convert wanted quantity from moles to grams.

Step One: Convert grams to moles

3.5 grams * 1 mole/2.02 grams
1.7 moles

Step Two: Multiply my answer by the molar ratio

1.7/1 * 2/3

1.1 moles

Step Three: Convert from moles to grams

Is this where I'm stumbling or did I mess up earlier?

 

[Math Is Hard]

I'm afraid that method loses me. Maybe someone else can help.

But I'll give you the rest of my thoughts for what it's worth..

to finish the problem, there is a number that you can multiply to the 0.1776g of hydrogen to make it 3.5 g. Find that. Then multiply that same number to the 1 g of ammonia.

for 3.5 g of hydrogen, I end up with about 19.7 g of ammonia.

 

[AngelShare]

I hope someone else will come along but, at this point, it doesn't seem likely...I don't think anyone else has come around since the first page.

Thanks for the help, I really appreciate your coming back to help me even though this thread has gone on for about eight days now.

 

[quantumdude]

OK, I'm listening. Can you summarize the open problems so I don't have to go through the whole thread?

Thanks,

 

[AngelShare]

Copied and pasted:

Problem #2:
Determine the number of grams of NH3 produced by the reaction of 3.5g of hydrogen gas with sufficient nitrogen gas. (Check for diatomic elements). Use the flowchart below for help.

Given quantity in grams--> change to moles----> multiply by ratio from balanced equation -----> convert wanted quantity from moles to grams.

Is this right:

Step One: Convert grams to moles

3.5 grams * 1 mole/2.02 grams
1.7 moles

Step Two: Multiply my answer by the molar ratio

1.7/1 * 2/3

1.133333333 moles

Step Three: Convert from moles to grams

1.133333333 moles/1 * 17.031/1 moles

19. 3018 grams

 

[ksinclair13]

Yes, that is close. The answer I got was 20. grams. Watch your significant figures. I learned not to round until the end for problems like these. But, your final answer should only have two regardless (either 19 or 20.).

I'll show you how you could have set it up in dimensional analysis:

3.5 g H_2(\frac{1 mol H_2}{2.02 g H_2})(\frac{2 mol NH_3}{3 mol H_2})(\frac{17.04 g NH_3}{1 mol NH_3}) = 20. g NH_3

I hope this doesn't confuse you more.

 

[AngelShare]

It doesn't confuse me, I can see similarities between your problem and mine. ^_^

 

[quantumdude]

Copied and pasted:

Problem #2:
Determine the number of grams of NH3 produced by the reaction of 3.5g of hydrogen gas with sufficient nitrogen gas. (Check for diatomic elements). Use the flowchart below for help.

Given quantity in grams--> change to moles----> multiply by ratio from balanced equation -----> convert wanted quantity from moles to grams.

Is this right:

Step One: Convert grams to moles

3.5 grams * 1 mole/2.02 grams
1.7 moles

Step Two: Multiply my answer by the molar ratio

1.7/1 * 2/3

1.133333333 moles

Step Three: Convert from moles to grams

1.133333333 moles/1 * 17.031/1 moles

19. 3018 grams

I haven't checked the numerical answer because I don't have a calculator handy, but your procedure is correct.

 

[AngelShare]

I turned it in and got a perfect grade, it was right! It may have taken me a lifetime to get it done but, "Give yourself a pat on the back! You are one of the few that got this lesson 100% - even on the second try!" definitely makes it completely worth it!

 

[chemgirl]

Problem #1
Aluminum oxide is formed from the reaction of metallic aluminum with oxygen gas. How many moles of Aluminum are needed to form 3.4 moles of Aluminum oxide?
Hint: This is a simple ratio problem. Just use the molar ratio from the balanced equation. This was covered in 5.09. Mass is not involved.

what is the answer to number one?? i get number 2, i just have no clue what to do with #1.
the balanced equatoin is Al + O2--> Al2O3?

 

[Kushal]

balance the equation first. from this obtain how many moles of Al reacts with O2.
and then use some logic to get to the answer.