chemistry
Zenah A
Worksheet8Chapter9-MoleRatioandLimitingReagent1.docxWorksheet 9 – Chapter 9
Mole Ratio and Limiting Reactants
Please do problems that are highlighted in yellow. You must show your work for credit.
Limiting reactants:
The first thing to know about limiting reactant situations is that there is nothing special about limiting reactant situations. Virtually every reaction that takes place involves a reactant that runs out while more of the other reactant(s) still remains.
Take baking chocolate chip cookies as an example. One recipe for CCC’s calls for the following ingredients: butter, sugar, brown sugar, vanilla extract, eggs, flour, baking soda, salt, and chocolate chips. You check your kitchen and, realizing that you have all of those ingredients, decide to make as many cookies as you can with what you have on hand.
Now think about this; how likely is it that you have exactly the right amount of every ingredient to make your C3 and that you will use up the last of each ingredient at exactly the same time? “Sweet, I have exactly the amount of everything I need to make 15-dozen cookies with no ingredient left over!” Um… no. More likely “Drat! I ran out of chocolate chips after making 8 cookies and now I can’t make any more! What am I going to do with all of these left-over ingredients?!?!”
The point? In virtually every situation where you are mixing ingredients, you are going to run out of one or more of those ingredients while also having other of those ingredients left over. This is the same in chemistry, though we call our ingredients “reactants” or “reagents”.
So, what does this mean to you as a chemist? Not too much actually, because there is no change to the chemistry that you have learned. You are changing amounts of reactants to amounts of products, so you will need to use mole-to-mole ratios. You will need to convert to and from moles in order to use the mole-to-mole ratio, so you will need things like molar mass, Avogadro’s Number, and other conversions involving the mole. Nothing new! The only “new’ thing here is that you are going to run out of one reactant during a reaction, and when you do, the reaction stops dead.
You already know everything that you need for this! Awesome! You just have to do it multiple times. Not so awesome!
Let’s start off by figuring out how to identify a limiting reactant problem in the first place. Without this skill, you are likely do too much (or too little) work. It turns out that, for reasons discussed above, most calculations that involve mole-to-mole ratios are limiting reactant situations. You will identify these situations as a limiting reactant problems when the problem gives you an amount of more than one reactant and asks you to calculate how much product will be produced.
Example:
What mass of PbCl2 will be made when 15.9 g of CaCl2 is added to a solution of Pb(NO3)2?
This is NOT treated as a limiting reactant problem because you are given an amount of only a single reactant (CaCl2). It actually is technically still a limiting reactant situation, but because you were not given an amount of the other reactant (Pb(NO3)2), you get to assume that you have an infinite amount of it and so will never run out of it. This means that the one reactant for which you were given an amount will run out and therefor limit how much product can be made.
Example:
What mass of PbCl2 will be made when 15.9 g of CaCl2 is added to a solution containing 5.13 g Pb(NO3)2?
This IS a limiting reactant problem because you are given an amount of more than one reactant (CaCl2 and Pb(NO3)2) and asked for an amount of product (PbCl2) formed. In the previous example, we could assume an infinite mass of Pb(NO3)2 because we were not given information about how much we has used. In this case, were are given specific amounts of each reactant, so that assumption is invalid. Now we have to figure out which reactant we are going to run out of rather than just knowing it.
It bears repeating that all chemical reactions involve a limiting reactant, even if they don’t look like it. In the first example above, we did not treat it as a limiting reactant problem because we assumed that there was infinite Pb(NO3)2 which meant that the 15.9 g of CaCl2 was automatically going to run out first. It is still a limiting reactant problem, we just didn’t need to figure out what the limiting reactant was.
If the question does NOT need to be treated as a limiting reactant problem, you simply convert the amount of the stuff that you were given to the amount of the stuff you asked for. (What do you ALWAYS have to use when going from an amount of one substance to an amount of another substance?) If it DOES need to be treated as a limiting reactant question, it is the exact same calculation, but you will need to do it more than once in order to identify the reactant that limits the reaction!
In a limiting reactant problem, you have set amounts of reactants, one of which you are likely to run out of while some of the other reactant remains.
Example:
Hydrogen gas and oxygen gas are mixed to make water. If 2.80 grams of hydrogen gas is reacted with 17.7 grams of oxygen gas, how much water can be formed and how much of the reactants will be left over?
The first step, as usual, is to get a balanced equation (you must do this for any of these problems), so:
The second step is to determine which reactant is limiting the amount of product that can be formed (i.e. the limiting reactant.) This is done by calculating how much of a single product can be formed from each of the given amounts of reactants. You must always compare the amount of the same product formed from each reactant or the comparison is invalid.
Start by determining how much water can be formed from the 2.80 g of H2. You are assuming here that you have enough oxygen to completely react with all of the hydrogen. This may or may not be a good assumption, but it is an assumption you must make in order to solve the problem.
Now determine how much water can be formed from the 17.8 g of O2. This time you are assuming that you have enough hydrogen to completely react with all of the oxygen. Again, this may or may not be a good assumption, but it is still an assumption you must make in order to solve the problem.
So, you can use up all of your H2 and make 1.37 mol of H2O or you can use up all of your O2 and make 1.11 mol of H2O. You only have 17.8 g of O2, so you cannot make any more H2O than the 1.11 mol shown in the second calculation. You ran out of O2 before you ran out H2, so the O2 was the limiting reactant.
From above, we have found that you can make 1.11 mol of H2O (O2 was limiting) from the given amounts of H2 and O2. This means that the theoretical yield of H2O is 20.0 g. The next part of the question is how much of each reactant is left after the reaction is complete.
Finding how much of the limiting reactant is left is a simple as asking yourself “How much of something is left when I run out of it?” The answer? None! By definition, if you run out of something, you have none of it left which means that the remaining amount of limiting reactant is always zero!
There are two ways to find out how much of the excess reactant is left. In one, you must first find out how much of the excess reactant was used in the reaction. You will do this by starting with the amount of the limiting reactant you were given in the question (17.8 g of O2) and converting it to the excess reactant (H2)
This is the amount of the H2 that was USED to make the H2O!
Be careful! 2.24 g H2 is not what is left over, it is the mass of hydrogen that was used up in making the water. You MUST subtract this amount from what you started with to find out how much is LEFT!
2.80 g H2 (started) – 2.24 H2 (used) = 0.56 g H2 LEFT OVER AFTER THE REACTION!
The second way (you can choose your favorite approach since they will give virtually the same answer) is to remember the “Law of Conservation of Mass”. You must start and end with the same mass! In other words, the total mass of reactants you had before the reaction started must be equal to the total mass of all products AND remaining reactants after the reaction is complete.
17.7 g O2 + 2.80 g H2 = 20.5 g total reactants before the reaction
20.0 g H2O + 0 g O2 + X g H2 = 20.5 g total after the reaction
X = 0.5 g H2 left after the reaction!
The difference between the two methods (0.56 g vs. 0.5 g) is due solely to sig. figs.
This was fairly long-winded, but hopefully helpful. The next part will be a more condensed version of “How to do limiting reactant problems”
Once you have identified a problem as a limiting reactant problem, here are the steps to go through:
Step 1: Write the balanced chemical equation
Step 2: Pick a product and find out what amount of that product can be made from each of the given reactants. You must always compare the amount the SAME product made from each reactant . If the reaction is A + B C + D + E, then you must pick ONE product, let’s say E in this case, and find out how much E can be made from A and how much E can be made from B . The smallest of the calculated product masses is how much of the product can actually be made (this is the theoretical yield)! The reactant leading to the smallest amount of product is the limiting reactant. Remember, all of the limiting reactant is used up, so the amount remaining is ALWAYS zero.
The following only needs to be done if the question asks for masses of the other products and reactants
Step 3: For the rest of the calculations, start with the amount of the limiting reactant given in the question. Calculate the masses of other products made from the given amount of the limiting reactant (since these are products, they are the amount MADE and therefore no subtraction is needed).
Step 4: Calculate the masses of the excess reagent(s) USED starting with the amount of the amount of the limiting reagent given. Reagents are always USED, so once you have calculated the amount used, you MUST subtract it from the amount of that reactant you were given.
I recommend making a table like I showed in class to keep track of everything!
Example 1: 4.529 g of antimony (III) sulfate and 13.98 g of lead (II) chlorate are reacted together. How much of all products and reactants are left after the reaction is complete?
This IS a limiting reactant question because amounts of the two reactants are given.
Step 1: balance equation:
Sb2(SO4)3(aq) + 3 Pb(ClO3)2(aq) 2 Sb(ClO3)3(aq) + 3 PbSO4(s)
Step 2: I am going to pick the lead (II) sulfate as my product just because I can. It doesn’t matter what product you start with, you will get the same answers!
7.750 g PbSO4 is the smallest amount of product calculated, so that is how much can actually be made. This also shows that Sb2(SO4)3 is the limiting reactant (of which there will be none left).
|
Sb2(SO4)3 = 0 g |
Sb(ClO3)3 |
|
Pb(ClO3)2 |
Pb(SO4)2 = 7.750 g |
Step 3:
This is the amount of the second product that can actually be made!
|
Sb2(SO4)3 = 0 g |
Sb(ClO3)3 = 6.339 g |
|
Pb(ClO3)2 |
Pb(SO4)2 = 7.750 g |
Step 4:
This is the amount of the excess reactant that was USED!!
13.98 g Pb(ClO3)2 (started) – 9.56006 g Pb(ClO3)2 (used) = 4.42 g Pb(ClO3)2 LEFT!!
This last step is the step the most people forget, so DON’T forget!
OR
(started masses) – (final masses) = excess reactant left
(4.529 + 13.98) – (6.339 + 7.750) = 4.42 g excess left
Either way, the final answer is:
|
Sb2(SO4)3 = 0 g |
Sb(ClO3)3 = 6.339 g |
|
Pb(ClO3)2 = 4.42 g |
Pb(SO4)2 = 7.750 g |
Example 2: What is the mass of BOTH products after the reaction between 1.14g of potassium carbonate with 1.15g of tin (IV) nitrate?
2 K2CO3(aq) + Sn(NO3)4(aq) Sn(CO3)2(s) + 4 KNO3(aq)
0.815 g Sn(CO3)2 can be made and Sn(NO3)4 is the L.R (none left).
Once you have run out of one of the reactants, you CANNOT make ANY more of ANY of the products. This means that once you have identified the limiting reactant for one product, the same reactant will be the limiting reactant for ALL PRODUCTS!!! The only reason you had to do 2 calculations (one for each of the reactants) above was because you had to identify the limiting reactant. Now that you know what the L.R. is, you only need to do one calculation for each remaining products starting with the given amount of the L.R.
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Sn(CO3)2 = 0.814 g made |
|
KNO3 = 1.38 g made |
You try:
1) Calculate the masses of ALL reactants and the theoretical yield of all products when 25.00 g of nickel (II) chloride is reacted with 25.00 g of silver sulfate
2) Calculate the theoretical yield of all products when 18.9 grams of lithium phosphate reacts with excess gold (I) acetate
3) A vessel contains 0.611 grams of butane gas (C4H10(g)) and 3.651x1022 molecules of oxygen gas. After the butane is reacted with the oxygen, how many molecules of CO2(g) are formed and what is the theoretical yield of water?
4) Calculate the masses of ALL reactants and the theoretical yield of all products when 0.500 grams of copper (II) nitrate is reacted with 1.25 grams of sodium oxalate
5) Calculate the theoretical yield of strontium sulfate when 525 mL of 1.199 M sulfuric acid is added to 23.66 grams of strontium bromide
Dealing with moles:
Avogadro’s number is the key to this type of problem. Everyone knows what is meant when someone says they have a dozen somethings. It is automatic that a dozen means there are 12 of whatever somethings are being talked about. Avogadro’s number is the same thing except a little bigger and is used to describe the number of particles in one mole of substance. If I told you that I had a dozen pencils and asked you to tell me how many pencils I have, this is what you would do to find out:
You can also find how many dozen pencils you have if you have 60 pencils:
Just as 1 dozen pencils contains 12 pencils, 1 mole of particles contains 6.022x1023 particles. These particles can be anything, but in this class, they will always be atoms, molecules, or formula units. A problem asking how many atoms there are in a certain number of moles will be set up EXACTLY the same way as the proceeding example of 3 dozen pencils.
Examples:
a) How many atoms are there in 5.62 moles of potassium?
b) How many moles are there in 1.16x1022 molecules of Br2?
Now for a little twist. These will be basically the same questions with one added step. In this case, you will be asked how many atoms or moles are represented by one PART of a molecule.
Examples:
a) How many sodium atoms are there in 3.75 moles of Na3PO4?
The first step is to determine the number of sodium moles in one mole of Na3PO4. Look at the subscript to find out. There are 3 sodium moles in each mole of the compound. So:
b) How many sulfur atoms are there in .512 moles of SO2?
Once again, how many moles of sulfur are there in each mole of SO2? Looking at the subscript, we see that there is one mole of sulfur for each mole of SO2. So:
c) If there are 7.84x1024 atoms of oxygen in a sample of hydrogen peroxide (H2O2), then how many moles of hydrogen peroxide are present?
This is the last question, only in reverse. First find the number of moles of oxygen atoms in each mole of H2O2, which from the subscript, is 2. So:
NOTE: The number of moles of one substance over the number of moles of another substance is called the mole to mole ratio. In the preceding example, the mole to mole ratio is 1:2, meaning there is one mole of H2O2 for every two moles of O. The mole to mole ratio is used in the majority (but not all) of the chemical calculations you will do in this class, so learn it well!
d) How many moles of magnesium nitrate are present in a sample containing 1.14x1025 atoms of oxygen?
First, find the mole to mole ration for oxygen to magnesium nitrate. There are 6 mole of O for each one mole of Mg(NO3)2 so the mole to mole ratio is 6 to 1. How did I know this? Look at the subscript OUTSIDE the parentheses, it is a 2. Now multiply that but the number of oxygen INSIDE the parentheses (which is 3) and you get a TOTAL of 6 oxygen for each Mg(NO3)2. So:
That’s right… your turn yet again. Think the problems through. They are all done the same way, but don’t just go through the motions, think about them and understand them!
1) If a 1 liter flask contains 55.4 moles of H2O, how many atoms of hydrogen are in the flask?
2) A sample has 1.93x1020 atoms of Iron. How many moles of Fe2O3 would this make?
3) How many atoms of carbon are there in 6.15x10–8 moles of (CH3CH2CO2)2C6H4
4) How many moles of Ra(CN)2 are present if you have 8.34x1027 atoms of N
5) How many atoms of hydrogen are there in 1.23x10–4 moles of Aspartame, C14H18N2O5
6) Calculate the mass (in grams) of 4.97x1024 molecules of sulfur dioxide.
You have just gone over how to go back and forth between particles and moles. The next step is to start or end with a mass. The same methods will be used for this type of problem as were used in the last problems, but now a molecular or atomic mass will also be used. At this point, it may also help to start writing out a game plan for yourself before doing each problem. When you write a game plan, think about what is takes to get from one step to the next. If you are going from grams to moles, you will ALWAYS use the molar mass. Going from moles of one substance to moles of another will ALWAYS require a mole to mole ratio. Going from moles to grams will ALWAYS require a molar mass. If you write out a games plan for each problem, and now what each step requires, you are set to get the problem correct!
Examples:
a) How many atoms of oxygen are there in 52.6 g of (NH4)3PO4?
Game plan: grams (NH4)3PO4 mol (NH4)3PO4 mol O atoms O
b) How many grams of mercury (II) bromide are present if you have 8.52x1019 atoms of mercury?
Game plan: atoms Hg mol Hg mol HgBr2 grams HgBr2
c) What mass of fluorine is in 5.78x1024 formula units of NaBF4?
Game plan: formula units NaBF4 mol NaBF4 mol F grams NaBF4
Your turn …. Never mind, you know what to do.
1) What mass of rubidium is in 2.39x1020 molecules of Rb2SO4?
2) How many sodium atoms are there in 6.53kg of NaHCO3?
3) What is the mass of iodine in 8.52x1020 molecules of CeI3?
4) How many oxygen atoms are in 1.2x10–3 grams of testosterone, C19H28O2?
5) What is the mass of a snowflake (think) containing 2.67x1019 atoms of hydrogen?
6) How many nitrogen atoms in 1.05x1021 molecules of caffeine, C8H10N4O2?
7) What is the mass of a Vitamin C tablet that has 5.13x1021 atoms of oxygen? The formula for Vitamin C is C6H8O6.
8) A sample of calcium acetate contains 5.98x1023 atoms of oxygen. What mass of hydrogen does it contain?
9) Home many iodine atoms are there in 28.3mg of carbon tetraiodide?
