Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Dave82 on October 17, 2006, 06:31:50 AM
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EDITED YET AGAIN:
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..
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 = (89g / 142.7g) x 100 = 62.37%
Did i get the actual and theoretical values correct or switch them?
==============================
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?
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
Max mass = 45.06g NH3
% yield of NH33 = (actual yield / theoretical yield) x 100 = (41.6g / 45.06g) x 100 = 92.32%
correct?
==============================
N2 + 3 H2 ----> 2 NH3
Calculate the theoretical yield of ammonia when 8.9 grams of hydrogen react with 56.4
grams of nitrogen.
*** The fact that the amounts of reactants are given tells us that this is a limiting-reactant problem***
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
Find out which is the limiting reagent:
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 (1 mol NH3 / 17.034 g) = 0.1728g NH3
I"M STUCK HERE :(
===============================
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.
*** The fact that the amounts of reactants are given tells us that this is a limiting-reactant problem***
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
Find out which is the limiting reagent:
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 H2) = 2.123 mols NH3
Therefore:
N2 is the limiting reactant bc it yields fewer mols of NH3
Convert moles of NH3 to grams:
Mass of NH3 = 2.123 mols NH3 x (17.034 g NH3 / 1 mol NH3) = 36.17 g NH3
is that actual or theoretical??
=========================
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I've checked the second problem only.
mols of H2 = 8g H2 x (1 mol H2 / 1.016g H2) = 7.874 mol H2
MW of hydrogen gas is 2.016g/mol, which means you have 3.968 moles of H2.
And that, believe me, makes a difference when you calculate the yield. ;)
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I am afraid you are four times wrong.
% yield of NaNO3 = (actual yield / theoretical yield) x 100 =
(89g / 142.7g) x 100 = 62.37%
89g is not an actual yield.
(8.9g H2 + 56.4g N2 = 65.3g NH3)
No, this is a limiting reagent question. You are asked about product mass, not %.
Now convert moles of H2 to moles of NH3 (use molar ratio from balanced equation):
No, this again starts as a limiting reagent question, once you will find out limiting reagent you may calculate theoretical yield and use this number for % calculation.
All that, plus Albert is right as well.
Note: you don't have to go through conversion to moles each time. You may calculate everything using molar mass ratios based stoichiometry calculations (http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions).
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Given that I showed you what is wrong in the second question, http://www.chemicalforums.com/index.php?topic=10841.msg51156#msg51156, check this out in problem #1:
% yield of NaNO3 = (actual yield / theoretical yield) x 100 =
(89g / 142.7g) x 100 = 62.37%
Are you sure it's 89g and not 133.4g, instead? ;)
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did i flip them around?
is this correct?
% yield of NaNO3 = (actual yield / theoretical yield) x 100 =
(142.7g / 89g ) x 100 = 60.34%
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did i flip them around?
No, you have used wrong value. Reread the question.
(142.7g / 89g ) x 100 = 60.34%
Have you calculated it? :o
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well since it came over 100, i thought i'd subtract 100
obviously not correct.
I'm gonna go over the book and then re-post after i have done a lot of work.
Hopefully you will still be online Borek to check my work :D
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EDITED YET AGAIN:
Don't do that - better post again in the same thread. When it is edited it is not displayed as a new post, besides, Albert's and mine previous comments are taken out of context.
-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...
% yield of NaNO3 = (actual yield / theoretical yield) x 100 = (89g / 142.7g) x 100 = 62.37%
Did i get the actual and theoretical values correct or switch them?
Still wrong. Plese read the question carefully - ho much sodium nitrate was produced?
92.32%
This one looks OK, although I would use only two significant digits: 92%.
Mass (g) of NH3 = 2.943 mols NH3 x (1 mol NH3 / 17.034 g) = 0.1728g NH3
Close. Moles are OK, final conversion is screwed.
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 H2) = 2.123 mols NH3
Check your math.
Convert moles of NH3 to grams:
Mass of NH3 = 2.123 mols NH3 x (17.034 g NH3 / 1 mol NH3) = 36.17 g NH3
is that actual or theoretical??
None, as it is wrong - but if it were OK it would be theoretical ;)
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EDITED YET AGAIN:
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..
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 = 93.48% = 93%
:D
==============================
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?
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
Max mass = 45.06g NH3
% yield of NH3 = (actual yield / theoretical yield) x 100 = (41.6g / 45.06g) x 100 = 92.32% = 92%
==============================
N2 + 3 H2 ----> 2 NH3
Calculate the theoretical yield of ammonia when 8.9 grams of hydrogen react with 56.4
grams of nitrogen.
*** The fact that the amounts of reactants are given tells us that this is a limiting-reactant problem***
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
Find out which is the limiting reagent:
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) = 50.13g NH3
% yield of NH3 = (actual yield / theoretical yield) x 100 = (????g / 50.13g) x 100 =
===============================
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.
*** The fact that the amounts of reactants are given tells us that this is a limiting-reactant problem***
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
Find out which is the limiting reagent:
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 H2) = 2.123 mols NH3
Therefore:
N2 is the limiting reactant bc it yields fewer mols of NH3
Convert moles of NH3 to grams:
Mass of NH3 = 2.123 mols NH3 x (17.034 g NH3 / 1 mol NH3) = 36.16 g NH3
36.16g NH3 (1 mol N2 / 2 mol NH3) = 18.01g NH3
% yield of NH3 = (actual yield / theoretical yield) x 100 = (18.01g / 36.16g) x 100 = 50.00% = 50%
??
=========================
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Mass (g) of NH3 = 2.943 mols NH3 x (17.034 g / 1 mol NH3) = 50.13g NH3
Bingo. That's your final nswer.
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 H2) = 2.123 mols NH3
Check your math in quoted text.
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EDITED YET AGAIN:
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..
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 = 93.48% = 93%
:D
==============================
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?
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
Max mass = 45.06g NH3
% yield of NH3 = (actual yield / theoretical yield) x 100 = (41.6g / 45.06g) x 100 = 92.32% = 92% well technically 90% in sig figs
==============================
N2 + 3 H2 ----> 2 NH3
Calculate the theoretical yield of ammonia when 8.9 grams of hydrogen react with 56.4
grams of nitrogen.
*** The fact that the amounts of reactants are given tells us that this is a limiting-reactant problem***
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
Find out which is the limiting reagent:
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) = 50.13g NH3 = 50.g NH3
===============================
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.
*** The fact that the amounts of reactants are given tells us that this is a limiting-reactant problem***
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
Find out which is the limiting reagent:
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 / 3 mols H2) = 2.831 mols NH3
Therefore:
N2 is the limiting reactant bc it yields fewer mols of NH3
Convert moles of NH3 to grams:
Mass of NH3 = 2.831 mols NH3 x (17.034 g NH3 / 1 mol NH3) = 48.23 g NH3
36.16g NH3 (1 mol N2 / 2 mol NH3) = 18.01g NH3
% yield of NH3 = (actual yield / theoretical yield) x 100 = (18.01g / 48.23g) x 100 = 37.34% = 37%
=========================
AM I ALL GOOD ;D
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Don't quote those questions that are already OK.
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 / 3 mols H2) = 2.831 mols NH3
Still wrong. Two moles of ammonia are produced for every mole of nitrogen.
AM I ALL GOOD ;D
Still questionable ;)
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AHHHH, i dont know why i did that!!!
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.
*** The fact that the amounts of reactants are given tells us that this is a limiting-reactant problem***
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
Find out which is the limiting reagent:
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:
H2 is the limiting reactant bc it yields fewer mols of NH3
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 = (71.33g / 98.2g) x 100 = 72.64% = 73%
-
Mass of NH3 = 6.283 mols NH3 x (17.034 g NH3 / 1 mol NH3) = 107.0 g NH3
That's your theoretical yield; I don't know (and don't want to) what you did later...
-
I can see this is irritating you, so hopefully this is it.
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.
*** The fact that the amounts of reactants are given tells us that this is a limiting-reactant problem***
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
Find out which is the limiting reagent:
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:
H2 is the limiting reactant bc it yields fewer mols of NH3
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 = 91.78% = 92%
-
I can see this is irritating you, so hopefully this is it.
I am rather patient. But even Universe is not infinite ;)
% yield of NH3 = (actual yield / theoretical yield) x 100 = (98.2g / 107.0g) x 100 = 91.78% = 92%
OK