Limiting Reactants And Percent Yield? WHAT? Chem Help Needed

I have this paper about limiting reactants and reagents and what not, and there's about 25 problems, I just need help on a few that require you to calculate percent yeild. I can do the limiting reactants, but no idea how to do the percent yield.

[ba7cfc40c10]Any Help Would Be Greatly Appreciated!!!
Karma Will Be Given!!![/ba7cfc40c10][/sizea7cfc40c10]

2 CR(NO3)3 + 3 Na2CO3 --> 1 Cr2(CO3)3 + 6 NaNO3

-If 89 grams of sodium carbonate are reacted with an excess of chromium(III)nitrate and only 133.4 grams of sodium nitrate are collected, what is the percent yield......

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N2 + 3 H2 ----> 2 NH3

-What is the maximum mass of ammonia that can be obtained by reacting 8 grams of hydrogen with an excess of nitrogen? If only 41.6 grams of ammonia could be recovered, what was the percent yield?

-Calculate the theoretical yield of ammonia when 8.9 grams of hydrogen react with 56.4 grams of nitrogen.

-A sample 19 grams of hydrogen reacted with 119 grams of nitrogen producing 98.2 grams of ammonia. Calculate the theoretical and percent yield.

Heh, I remember doing that. I'm glad I'm done with it. It's really not that hard when you know what to do and how to do it. Too bad I forgot how. Anyway, I wish you the best of luck with that. There's gotta be someone here who remembers how to do it.

Is this regular chem or Ap chem?

yea i used to be really good at that sorta stuff too. As soon as i didn't really need to know it for class/tests my knowledge straightup vanished though, so um yea good luck

It's Chemistry II, a new class at our school....

Basically AP Chemistry I I guess.....

has this become the homework help forum.
i have a test tomorrow and my gf (who is great at chem) is helping with someone's econ homework. but if it isnt due tomorrow, let me know and i'll have it in your pm box tomorrow.

Yeah it's due tomorrow (

Thanks for the effort though...

-To anyone else, I still have no clue as to what I'm doing, any help is still greatly appreciated (

Read this
http//www.uwosh.edu/faculty_staff/kedrowsk/yield_calc.htm

I remember doing this buy I forgot how/ ( Its not that hard.

FROM THE WIFEY

SCROLL TO END FOR CORRECTED (well.... i think corrected
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if someone could check this, it would be great.

oh no wait, these are all wrong. Ill post for you before you class like.

[quote56cb6ac328]
2 Cr(NO3)3 + 3 Na2CO3 --> 1 Cr2(CO3)3 + 6 NaNO3

-If 89 grams of sodium carbonate are reacted with an excess of chromium(III)nitrate
and only 133.4 grams of sodium nitrate are collected, what is the percent yield......
[/quote56cb6ac328]

2 Cr(NO3)3 + 3 Na2CO3 --> 1 Cr2(CO3)3 + 6 NaNO3

89 g = sodium carbonate = Na2CO3 reacted

133.4 g = sodium nitrate = NaNO3 recovered


First convert the grams of Na2CO3 to mols
mols of Na2CO3 = 89g Na2CO3 x (1 mol Na2CO3 / 106.0g Na2CO3) = 0.8396 mol Na2CO3


Now convert moles of Na2CO3 to moles of NaNO3 (use molar ratio from balanced equation)
0.8396 mol Na2CO3 x (6 mols NaNO3 / 3 mols Na2CO3) = 1.679 mols NaNO3


Convert moles of NaNO3 to grams
Mass of NaNO3 = 1.679 mols NaNO3 x (85.00 g NaNO3 / 1 mol NaNO3) = 142.7 g NaNO3


% yield of NaNO3 = (actual yield / theoretical yield) x 100 = (133.4g / 142.7g) x 100 = [b56cb6ac328]93.48% = 93%[/b56cb6ac328]





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[quote56cb6ac328]
N2 + 3 H2 ----> 2 NH3

What is the maximum mass of ammonia that can be obtained by reacting 8 grams of
hydrogen with an excess of nitrogen? If only 41.6 grams of ammonia could be
recovered, what was the percent yield? [/quote56cb6ac328]

N2 + 3 H2 ----> 2 NH3

8 g = hydrogen = H2 processed

41.6 g = ammonia = NH3 recovered


First convert the grams of H2 to moles
mols of H2 = 8g H2 x (1 mol H2 / 2.016g H2) = 3.968 mol H2


Now convert moles of H2 to moles of NH3 (use molar ratio from balanced equation)
3.968 mol H2 x (2 mols NH3 / 3 mols H2) = 2.646 mols NH3


Convert moles of NH3 to grams
Mass of NH3 = 2.646 mols NH3 x (17.034 g NH3 / 1 mol NH3) = 45.06 g NH3


[b56cb6ac328]Max mass = 45.06g NH3[/b56cb6ac328]

% yield of NH3 = (actual yield / theoretical yield) x 100 = (41.6g / 45.06g) x 100 = [b56cb6ac328]92.32% = 92%[/b56cb6ac328]
technically 90% if your teacher is picky about sig figs...






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[quote56cb6ac328]
N2 + 3 H2 ----> 2 NH3

Calculate the theoretical yield of ammonia when 8.9 grams of hydrogen react with 56.4
grams of nitrogen. [/quote56cb6ac328]

lilili The fact that the amounts of reactants are given tells us that this is a limiting-reactant problemlilili

N2 + 3 H2 ----> 2 NH3

8.9 g = hydrogen = H2
56.4 g = nitrogen = N2




First convert the grams to moles
mols of H2 = 8.9g H2 x (1 mol H2 / 2.016g H2) = 4.415 mol H2
mols of N2 = 56.4g N2 x (1 mol N2 / 28.02g N2) = 2.013 mol N2


[b56cb6ac328]Find out which is the limiting reagent[/b56cb6ac328]

Using H2
mols of H2 = 8.9g H2 x (1 mol H2 / 2.016g H2) = 4.415 mol H2
mols of NH3 (in this case) = 4.415 mol H2 x (2 mols NH3 / 3 mols H2) = 2.943 mols NH3

Using N2
mols of N2 = 56.4g N2 x (1 mol N2 / 28.02g N2) = 2.013 mol N2
mols of NH3 (in this case) = 2.013 mol N2 x (2 mols NH3 / 1 mols H2) = 4.026 mols NH3

Therefore
H2 is the limiting reactant bc it yields fewer mols of NH3



Convert from mols of NH3 to grams
Mass (g) of NH3 = 2.943 mols NH3 x (17.034 g / 1 mol NH3) = [b56cb6ac328]50.13g NH3 = 50.g NH3[/b56cb6ac328]




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[quote56cb6ac328]
N2 + 3 H2 ----> 2 NH3

A sample 19 grams of hydrogen reacted with 119 grams of nitrogen producing 98.2
grams of ammonia. Calculate the theoretical and percent yield. [/quote56cb6ac328]

lilili The fact that the amounts of reactants are given tells us that this is a limiting-reactant problemlilili

N2 + 3 H2 ----> 2 NH3

19 g = hydrogen = H2
119 g = nitrogen = N2

98.2 g = ammonia = NH3


First convert the grams to moles
mols of H2 = 19g H2 x (1 mol H2 / 1.016g H2) = 18.70 mol H2
mols of N2 = 119g N2 x (1 mol N2 / 28.02g N2) = 4.427 mol N2


[b56cb6ac328]Find out which is the limiting reagent[/b56cb6ac328]

Using H2
mols of H2 = 19g H2 x (1 mol H2 / 2.016g H2) = 9.425 mol H2
mols of NH3 (in this case) = 9.425 mol H2 x (2 mols NH3 / 3 mols H2) = 6.283 mols NH3

Using N2
mols of N2 = 119g N2 x (1 mol N2 / 28.02g N2) = 4.247 mol N2
mols of NH3 (in this case) = 4.247 mol N2 x (2 mols NH3 / 1 mols N2) = 8.494 mols NH3

Therefore
[u56cb6ac328]H2 is the limiting reactant bc it yields fewer mols of NH3[/u56cb6ac328]


Convert moles of NH3 to grams
Mass of NH3 = 6.283 mols NH3 x (17.034 g NH3 / 1 mol NH3) = 107.0 g NH3

107.0g NH3 (2 mols NH3 / 3 mols NH3) = 71.33g NH3


% yield of NH3 = (actual yield / theoretical yield) x 100 = (98.2g / 107.0g) x 100 =[b56cb6ac328] 91.78% = 92%[/b56cb6ac328]


=========================

Alright, i think this is all okay

oh wow I remember those days of chem.....(last semester). Never again.

did you ever use this?

if u still need help i can help. u the percent yeild the the actual yeild (calculated using the limiting reactant) divided by the theoritical yeild amount said to be produced in the problem

[quotebd576106c5="Dave82"]did you ever use this?[/quotebd576106c5]

Yes sir, and thanks to you and your awesome girlfriend, I got an 87 on the test, and I'm very pleased with it. It would have been a 91 but I forgot I had skipped a problem which I never went back and did. Thanks a lot!

[quotead1f8c11f6="johnjimjones"]oh wow I remember those days of chem.....(last semester). Never again.[/quotead1f8c11f6]

same here. I got a crapass 430-something on the California Standardized Testing thing but considering how the only thing i remember from Chem are the Gas Laws (Boyle' and such), it might not be half bad

[quoted3369c821f="Aurelius"][quoted3369c821f="johnjimjones"]oh wow I remember those days of chem.....(last semester). Never again.[/quoted3369c821f]

same here. I got a crapass 430-something on the California Standardized Testing thing but considering how the only thing i remember from Chem are the Gas Laws (Boyle' and such), it might not be half bad[/quoted3369c821f]


Heh we're doing that now, and yeah it is pretty easy as long as you memorize the formulas and such

haha aww, you should have posted sooner. She asked me the next day how that kid did on his test. She was genuinely concerned and all i could say was, "i dont know. pass the salt"