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

[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.