Stoichiometry Practice Problems

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  • #26
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1. 6.8 moles

Aw God, number two is gonna kill me.:cry: 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.:mad: 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.:uhh: ) 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.:grumpy:

2. 11.9 grams

How 'bout that?
 
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  • #27
Math Is Hard
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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.:redface:

Sorry, I have not had a chance to look at your second problem yet.
 
  • #28
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:rofl: I know, I could feel my sanity slipping there.:tongue:

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.:uhh: 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" [Broken] and used it.:wink: )

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.:rofl: 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?:bugeye: 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.:redface: 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?:surprised

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.:cry:

Just had to rant a bit there,:wink: 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:
 
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  • #29
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.
 
  • #30
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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).
 
  • #31
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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 wanna check my math on this? It's been a while since I've done these.
 
  • #32
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You tried it too, Math Is Hard? I'm still working on it as well.:rofl:

#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.:biggrin:
 
  • #33
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how would you calculate the molar mass of 3H2?
how would you calculate the molar mass of 2NH3?

try that first.
 
  • #34
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3H2 = 6.048
2NH3 = 34.062

Is that right?
 
  • #35
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AngelShare said:
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?
 
  • #36
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Math Is Hard said:
now what?

That's a good question...:rofl:

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?
 
  • #37
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I'm afraid that method loses me. :frown: 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.
 
  • #38
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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.:rofl:

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.:smile: :tongue:
 
  • #39
Tom Mattson
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OK, I'm listening. Can you summarize the open problems so I don't have to go through the whole thread?

Thanks,
 
  • #40
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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
 
  • #41
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:

[tex]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[/tex]

I hope this doesn't confuse you more.
 
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  • #42
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It doesn't confuse me, I can see similarities between your problem and mine. ^_^
 
  • #43
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AngelShare said:
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.
 
  • #44
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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!:biggrin:
 
  • #45
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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???
 
  • #46
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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.
 

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