Great job studying and doing your best to master limiting reagents!


Summary  all explanations in one!
POP QUIZ: CHALLENGE PROBLEM
For the balanced equation shown below, if the reaction of 1,567 grams of C4H8O and 5,792 grams of O2 produces a 61.5% yield, how many grams of CO2 would be produced?
2C4H8O+11O2=>8CO2+8H2O
For the balanced equation shown below, if the reaction of 1,567 grams of C4H8O and 5,792 grams of O2 produces a 61.5% yield, how many grams of CO2 would be produced?
2C4H8O+11O2=>8CO2+8H2O
Just in case you forgot or didn't understand any of the material, here is the most essential information from each section:
How to Find Limiting Reagents:
Limiting Reagent  a chemical that runs out first.
Moles – the mole is the measure of the amount of chemical substance.
The abbreviation for Moles is mol.
Molar mass [(Gram Formula Weights)(GFW)]  How much each element weighs.
e.g. Carbon (C) has a GFW of 12.
The unit for GFW is g/mol or grams per mole.
Here is a link to a great online Periodic Table of Elements to help you find the GFW.
http://www.webelements.com/
Look at each element's mass (the number under the symbol for the element). Round the element's weight to the nearest whole number. (except for Cl  keep at 35.5g/mol)
How to find GFW of a compound:
Add the weights of the elements together:
Compound: CaCO3
Ca x 1, C x 1, O x 3
Ca: 40 x 1 = 40
C: 12 x 1 = 12
O: 16 x 3 = 48
40 + 12 + 48 = 100 g/mol
Equations you will need:
# of moles = # of grams (g) / GFW
How to Use Limiting Reagents:
Step 1: Find the limiting reagent:
For the balanced equation shown below, if 96.1 grams of C2H3F were reacted with 142 grams of O2, how many grams of CO would be produced?
2C2H3F+3O2=>4CO+2H2O+2HF
C2H3F: 96.1 grams / 46 grams/mole = 2.089 moles (3) = 6.267391304347827 moles
O2: 142 grams / 32 grams/mole = 4.4375 moles (2) = 8.875 moles
6.267 moles < 8.875 moles
C2H3F is the limiting reagent.
Step 2: Compare the coefficients:
In the equation, look at the coefficient of the limiting reagent and of the compound for which you want to find the number of grams that will be produced by the reaction. They are expressed as:
2C2H3F and 4CO
Compare the coefficients through a ratio.
2 : 4 (simplified to) > 1 : 2
If you have a much more complicated ratio, such as 13 : 4 , then you can reduce the ratio through decimals. 13/13 = 1, 4/13 = 0.308
To find the number of moles for CO, multiply the decimal you got from the ratio by the original number of moles from C2H3F: (0.308)(2.089) = 0.643 moles (for CO).
Step 3: Find the number of grams for the desired compound:
Now we have already found how many moles the limiting reagent has. To find the amount of moles for the desired compound, use the ratio. Once you have found the amount of moles for the desired compound, use the equation: Moles = Grams / GFW
Multiply the amount of moles by the GFW of the compound to find the amount of grams.
2.089 moles (2) = 4.178 moles
4.178 (28) = 117 grams
How to Calculate the Percent Yield:
Equation:
% Yield = ( Actual yield / theoretical yield ) x 100
In this type of problem, the phrase “produces” denotes the actual yield and all you have to do is calculate the theoretical yield and plug in the numbers to achieve the percent yield. Also, the limiting reagent in these questions is the reactant that the question mentions. It is just like Using Limiting Reagents, but with an extra step at the end.
There are two types of problems for percent yield. One question gives you the actual yield and asks for the percent yield, and the other gives you the percent yield and asks for the actual yield:
2 Worked Example Problems:
For the balanced equation shown below, if the reaction of 33.5 grams of H2O produces 53.9 grams of O2, what is the percent yield?
3CO2+4H2O=>C3H8+5O2
Step 1:
To answer this question, you use the mass ratio from the balanced equation to determine the theoretical yield.
Step 2:
theoretical yield of O2:
(mass of O2)/(mass of H2O)*given mass
160 grams per mole / 72.08 grams per mole * 33.5 grams = 74.4%
Step 3:
Then, take the percentage you just found:
(actual/theory)*100=%yield
(53.9/74.4)*100=72.4%
For the balanced equation shown below, if the reaction of 24.2 grams of C6H6O2 produces a 31.9% yield, how many grams of H2O would be produced ?
2C6H6O2+13O2=>12CO2+6H2O
Step 1:
To answer this question, you use the mass ratio from the balanced equation to determine the theoretical yield.
Step 2:
theoretical yield of H2O:
(mass of H2O)/(mass of C6H6O2)*given mass
(108.12) / (220.2)*24.2 = 11.9
Step 3:
Multiplying the theoretical yield by the percent (and dividing by 100), provides the actual yield.
(theory* % yield )/ 100 = actual yield
(11.9*31.9)/100 = 3.80 grams
Bibliography:
http://getrealgurlsnutrition.files.wordpress.com/2013/04/sushi61.jpg
http://thomasoliversnyc.com/images/site/asianfusion.gif
http://wps.prenhall.com/wps/media/objects/3310/3390270/imag0307/AAAUAUT0.JPGhttp://www.reactiongifs.com/wpcontent/uploads/2013/01/applause.gif
http://25.media.tumblr.com/30a93aeca36783d8256e43fd98c164f3/tumblr_mky2d6Jxdn1qj8jkvo1_500.gif
http://wiki.openstreetmap.org/wiki/File:Symbol_OK.svg
http://www.perfectduluthday.com/wpcontent/uploads/2011/10/sushi.jpg
http://theunsoldhome.com/wpcontent/uploads/2011/05/00384832.jpg
Sources:
http://science.widener.edu/svb/tutorial/percentyieldcsn7.html
http://science.widener.edu/svb/tutorial/limitreagentcsn7.html
http://science.widener.edu/svb/tutorial/usinglimitreagentcsn7.html
http://www.webelements.com/
How to Find Limiting Reagents:
Limiting Reagent  a chemical that runs out first.
Moles – the mole is the measure of the amount of chemical substance.
The abbreviation for Moles is mol.
Molar mass [(Gram Formula Weights)(GFW)]  How much each element weighs.
e.g. Carbon (C) has a GFW of 12.
The unit for GFW is g/mol or grams per mole.
Here is a link to a great online Periodic Table of Elements to help you find the GFW.
http://www.webelements.com/
Look at each element's mass (the number under the symbol for the element). Round the element's weight to the nearest whole number. (except for Cl  keep at 35.5g/mol)
How to find GFW of a compound:
Add the weights of the elements together:
Compound: CaCO3
Ca x 1, C x 1, O x 3
Ca: 40 x 1 = 40
C: 12 x 1 = 12
O: 16 x 3 = 48
40 + 12 + 48 = 100 g/mol
Equations you will need:
# of moles = # of grams (g) / GFW
How to Use Limiting Reagents:
Step 1: Find the limiting reagent:
For the balanced equation shown below, if 96.1 grams of C2H3F were reacted with 142 grams of O2, how many grams of CO would be produced?
2C2H3F+3O2=>4CO+2H2O+2HF
C2H3F: 96.1 grams / 46 grams/mole = 2.089 moles (3) = 6.267391304347827 moles
O2: 142 grams / 32 grams/mole = 4.4375 moles (2) = 8.875 moles
6.267 moles < 8.875 moles
C2H3F is the limiting reagent.
Step 2: Compare the coefficients:
In the equation, look at the coefficient of the limiting reagent and of the compound for which you want to find the number of grams that will be produced by the reaction. They are expressed as:
2C2H3F and 4CO
Compare the coefficients through a ratio.
2 : 4 (simplified to) > 1 : 2
If you have a much more complicated ratio, such as 13 : 4 , then you can reduce the ratio through decimals. 13/13 = 1, 4/13 = 0.308
To find the number of moles for CO, multiply the decimal you got from the ratio by the original number of moles from C2H3F: (0.308)(2.089) = 0.643 moles (for CO).
Step 3: Find the number of grams for the desired compound:
Now we have already found how many moles the limiting reagent has. To find the amount of moles for the desired compound, use the ratio. Once you have found the amount of moles for the desired compound, use the equation: Moles = Grams / GFW
Multiply the amount of moles by the GFW of the compound to find the amount of grams.
2.089 moles (2) = 4.178 moles
4.178 (28) = 117 grams
How to Calculate the Percent Yield:
Equation:
% Yield = ( Actual yield / theoretical yield ) x 100
In this type of problem, the phrase “produces” denotes the actual yield and all you have to do is calculate the theoretical yield and plug in the numbers to achieve the percent yield. Also, the limiting reagent in these questions is the reactant that the question mentions. It is just like Using Limiting Reagents, but with an extra step at the end.
There are two types of problems for percent yield. One question gives you the actual yield and asks for the percent yield, and the other gives you the percent yield and asks for the actual yield:
2 Worked Example Problems:
For the balanced equation shown below, if the reaction of 33.5 grams of H2O produces 53.9 grams of O2, what is the percent yield?
3CO2+4H2O=>C3H8+5O2
Step 1:
To answer this question, you use the mass ratio from the balanced equation to determine the theoretical yield.
Step 2:
theoretical yield of O2:
(mass of O2)/(mass of H2O)*given mass
160 grams per mole / 72.08 grams per mole * 33.5 grams = 74.4%
Step 3:
Then, take the percentage you just found:
(actual/theory)*100=%yield
(53.9/74.4)*100=72.4%
For the balanced equation shown below, if the reaction of 24.2 grams of C6H6O2 produces a 31.9% yield, how many grams of H2O would be produced ?
2C6H6O2+13O2=>12CO2+6H2O
Step 1:
To answer this question, you use the mass ratio from the balanced equation to determine the theoretical yield.
Step 2:
theoretical yield of H2O:
(mass of H2O)/(mass of C6H6O2)*given mass
(108.12) / (220.2)*24.2 = 11.9
Step 3:
Multiplying the theoretical yield by the percent (and dividing by 100), provides the actual yield.
(theory* % yield )/ 100 = actual yield
(11.9*31.9)/100 = 3.80 grams
Bibliography:
http://getrealgurlsnutrition.files.wordpress.com/2013/04/sushi61.jpg
http://thomasoliversnyc.com/images/site/asianfusion.gif
http://wps.prenhall.com/wps/media/objects/3310/3390270/imag0307/AAAUAUT0.JPGhttp://www.reactiongifs.com/wpcontent/uploads/2013/01/applause.gif
http://25.media.tumblr.com/30a93aeca36783d8256e43fd98c164f3/tumblr_mky2d6Jxdn1qj8jkvo1_500.gif
http://wiki.openstreetmap.org/wiki/File:Symbol_OK.svg
http://www.perfectduluthday.com/wpcontent/uploads/2011/10/sushi.jpg
http://theunsoldhome.com/wpcontent/uploads/2011/05/00384832.jpg
Sources:
http://science.widener.edu/svb/tutorial/percentyieldcsn7.html
http://science.widener.edu/svb/tutorial/limitreagentcsn7.html
http://science.widener.edu/svb/tutorial/usinglimitreagentcsn7.html
http://www.webelements.com/