[Dave82]

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??



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

[Albert]

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. 

[Borek]

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.

[Albert]

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? 

[Dave82]

did i flip them around?

is this correct?
% yield of NaNO3 = (actual yield / theoretical yield) x 100 =
(142.7g / 89g ) x 100 = 60.34%

[Borek]

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? 

[Dave82]

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 

[Borek]

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 

[Dave82]

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%






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

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%


??

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

[Borek]

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.

[Dave82]

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%






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

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 

[Borek]

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 

Still questionable

[Dave82]

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%

[Borek]

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...

[Dave82]

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%