Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: unsavedhero on September 29, 2013, 04:44:08 PM
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4NH3 + 5O2 --> 4NO + 6H2O In a certain experiment 2.00g of NH3 reacts with 2.50g of O2
a what is the limiting reactant
b how many grams of NO and of H2O form?
c How many grams of the excess reactant remain after the limiting reactant is completely consumed?
d show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass
Mod edit: Your added attempt has been moved to your later post. Please do not edit posts after they have been replied to. Dan
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You must show you have attempted the question, this is a Forum Rule (http://www.chemicalforums.com/index.php?topic=65859.0).
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You must show you have attempted the question, this is a Forum Rule (http://www.chemicalforums.com/index.php?topic=65859.0).
Oh I apologize. Do I upload a picture or do I just type up the work? I'll type what I tried and put it in the original post
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You must show you have attempted the question, this is a Forum Rule (http://www.chemicalforums.com/index.php?topic=65859.0).
Oh I apologize. Do I upload a picture or do I just type up the work? I'll type what I tried and put it in the original post
It's usually better to type it up because it will be tidier and easier to follow than handwritten work.
It is also possible to upload images:
http://www.chemicalforums.com/index.php?topic=69802.0
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well i did my best :(. The work I did from last night is extremely messy so I had to reattempt it just now only to be confused even harder than I was last night. ugh :(. the difficulty of the questions in my h.w escalated extremely quickly
For (a) I chose NH3 because O2 is usually excess because its in the air. However maybe I am wrong because it may be a closed system idk.
for (b) I got 2.0g of NO and .82g of H2O
4NH3 = 68g + 160g 5O2 --> 4NO = 72g 6H2O = 108g
2.00g NH3/68 = .0294 2.50g O2/160 = .015625 i divided to see by how much the mass decreased from the molar mass of each molecule to the ones given which are 2.00g and 2.50g. then multiply.. -__- i honestly have no idea wtf I am doing. I tried last night but now im getting different answers. Problem is the text book doesnt give the answer so Im not sure. All the other questions seemed easy to me and I got them right idk why this particular question stumped me
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Anyone? Please I am desperate :'(
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Use stoichiometry. It helps.
1 mole = 6.022 x 1023 particles.
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well i know thats what I'm trying to do. this problem is in my stoicheometry chapter :(
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For (a) I chose NH3 because O2 is usually excess because its in the air.
This is not how you do the problem.
What does "limiting reagent" mean?
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Yeah I figured out how to get limiting reagent. so 2.00g/17 = .1176mol NH3 2.50g/32 = .078125mol O2
(.078125mol O2 ) (4NH3/5O2 ) = .0625 so for every .078125mol of O2 you need .0625mol of NH3 which would make O2 the limiting reagent because there is plenty of NH3 .1176mol to be exact but only .078125mol of O2 however I dont know how to do the rest :(
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Based on the limiting reagent you've identified, and the amount of this reagent you are told you start with, what is the amount of products you anticipate generating, assuming 100% conversion yield?
Hint: the stoichiometry in the balanced equation you are given will tell you the relative amount of products generated compared to the amount of reactants you start with, on a molar basis.
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omg I finally figured it out. Im literally gonna cry lmao :(!! its 1.875g of NO and 1.6875g of H2O However I am not sure if I am correct because the text book doesnt tell me so if someone can verify for me that'd be nice. I did it by multiplying the stoicheometric ratios by the moles of O2 (because O2 was the limiting reactant) and converting that into grams. so to find grams of NO formed I did .078125mol O2 × (4NO)/5molO2) x 30g/1mol NO = 1.875 and did the same thing to find water. but now for C I have to find how many grams of excess reactant remain after limiting reactant is completely consumed and I got .9367g of NH3 because .078125 mol of O2 need .0625 mol of NH3 to fully react. So I subtracted .0625 from the available amount of moles which is .1176 and got .0551mol NH3 left. and converted it into grams by doing .0551 x 17 = .9367g the # 17 being the molar mass of NH3. so is this all correct? what do you guys think? how am I supposed to answer (D)? how do I show that my calculations are consistent with the law of conservation of mass?
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its 1.875g of NO and 1.6875g of H2O
Yes, these numbers are correct. You might want to use a smaller number of significant digits (how many in the input data?).
how am I supposed to answer (D)? how do I show that my calculations are consistent with the law of conservation of mass?
Just check that total mass after the reaction equals total mass before reaction.
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so I just find the molar masses on both sides of the reaction and show that they're equal? thank you. :)
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Not molar masses, total mass of all substances present BEFORE reaction must be identical to total mass of all substances present AFTER the reaction.
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And remember, there will be some unreacted non-limiting reagent left over!
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1.875g + 1.675g = 3.55g
2.00g of NH3 reacts with 2.50g of O2
2.00g + 2.50g = 4.50g
3.55g does not equal 4.50g ;_;
*EDIT* never mind I got it. I just added the left over NH3 that didnt react which is equal to about .9367g and it equaled to about 4.50. thank you to everyone who helped. i appreciate it