A Practice Problem On Stoichiometry -- ANSWERS

A Practice Problem on Stoichiometry -- ANSWERSConsider the reaction represented by the equation below for all parts of this question: 2 BrCl33 Cl2 Br2(a) If 4 moles of BrCl3 reacts according to the equation, how many moles of Cl2 will be formed? How many moles of Br2will be formed?Answers: 6 mol Cl2 and 2 mol Br2Setups/Work:32 mol BrCl4 mol BrCl used12 mol BrCl4 mol BrCl used6 mol Cl2 mol Br(b) If 3.54 moles of BrCl3 reacts according to the equation, how many moles of Cl2 will be formed? How many moles of Br2will be formed?Answers: 5.31 mol Cl2 and 1.77 mol Br2Setups/Work:32 mol BrCl3.54 mol BrCl used3.54 mol BrCl used12 mol BrCl5.31 mol Cl1.77 mol Br(c) When any number of moles of BrCl3 reacts according to the equation, which of the following is correct?2 BrCl3 3 Cl2 Br2i) The number of moles of Cl2 formed is less than the moles of BrCl3 reacted.ii) The number of moles of Cl2 formed is two-thirds of the moles of BrCl3 reacted.iii) Three moles of Cl2 are formed.iv) All of the above.v) (i) and (ii) only.vi) none of the aboveExplanation: Since the coefficient of Cl2 is larger than that of BrCl3, (i) is wrong (i.e., the moles of Cl2formed must be larger than the moles of BrCl3 reacted). As such, (ii) must also be wrong (since 2/3 ofsomething is “smaller” than the something itself). (iii) may be attractive to some, but it makes no sense!How could the number of moles formed always be “3” no matter how many moles of BrCl3 react? It can’tthe amount of products made is proportional to the amount of reactant that reacts. The correct answerwould as follows:As the construction of the prior problems shows (see below), since 3 is the coefficient of Cl2 and 2 is thecoefficient of BrCl3, the amount of Cl2 formed must be 3/2 of the moles of BrCl3 that reacts:Let y represent the number of moles of BrCl3 that reacts, then:y mol BrCl used 32 mol BrCly3mol Cl2represents the moles of Cl2 formed when y mol of BrCl3 react

Example of a “Level 3” type stoichiometry problem fromhttp://web.mst.edu/ gbert/reactor/Areactor.htmlThe starting screen looks like:I clicked on the Level 3 button and then on “new reaction” and got:The top bar graphs and mol values indicate the initial state of a system that containsthe indicated amounts of four substances, A, B, C, and D.Click “run reaction” and you get:

Now you know what the final state of the system is AFTER reaction has occurred.You are asked to determine the (standard) balanced chemical equation that representsthe reaction that occurred. This means you need to determine who the reactant andproduct species are, as well as their coefficients such that the coefficients are in thesmallest whole number ratio.How can you do this? Well, first note that:1) the amounts of A and D have gone up,2) the amount of B has gone down,3) and the amount of C has not changedThis means that:1) A and D are being produced as reaction occurs. They are products2) B is being consumed (“used up”) as reaction occurs. It is a reactant3) C is neither a reactant nor a product (no net change. It won’t appear in thebalanced equation representing this reaction)To get the coefficients, you need to first calculate how many moles of each specieshas changed. Then you need to look at the ratio of those changes. One way to help“see” all of this in a compact and organized way is to use an ICF table (“I” stands forinitial, “C” stands for change, and “F” stands for final). For the current example, thetable would initially look like this (based on the values from the applet):

(values in moles)ABCDI (initial)4.13.28.41.39.70.48.49.7C (change in)F (final)As I hope you will recall from your prior chemistry course (when you had H’s and T’sin the calorimetry unit), the change in a variable, often expressed as a , is defined as“final” – “initial”.So A here (in moles) 9.7 mol – 4.1 mol 5.6 moles.Similarly, B is 0.4 – 3.2 -2.8 mol (note the negative sign!), C is 8.4 – 8.4 0 mol (its value did not change, right?) D is 9.7 – 1.3 8.4 molThe table now becomes:(values in moles)ABCDI (initial)4.13.28.41.3C (change in) 5.6-2.80 8.4F (final)9.70.48.49.7[Note that the signs are consistent with our earlier conclusion that A and D form andthus are products, and B decreases and is thus a reactant.]** It is the ratio of the “changes” that equals the ratio of the coefficients in thebalanced chemical equation, because that’s what the coefficients in abalanced equation represent—the ratio in which reactants react and productsform (in moles)!So, to reduce this ratio to a whole-number one, you can either do it by “inspection”, ordo what you did when you first learned how to get an empirical formula from “moles ofatoms” values—divide each value by the smallest (not including the zero, obviously).So here, divide 5.6, -2.8, and 8.4 by 2.8, and you’ll get 2, -1, and 3 as the coefficientsof A, B, and D respectively. Being careful to recall that B is the reactant and A and Dproducts, the balanced equation follows:B 2A 3D

If you input that into the program (see below), you’ll see that it says “RIGHT” in coloredletters to indicate you have entered the correct answer:Just for completeness, if you input an incorrect answer, you will get a “sorry” response,as well as a “tutor” button, which if clicked, will simply give you the correct answer (no“tutoring”.

SOLUTION TO PRELAB STOICHIOMETRY QUESTIONA container initially has 5.82 moles of N2O, 8.14 moles of O2, and 1.03 moles of N2O4 in a box. Assumethat the only reaction that can occur in this box is the one represented by the following balancedchemical equation:2 N2O(g) 3 O2(g) 2 N2O4(g)Suggestion: I would draw a picture showing what you start with (and then what you end up with in each case) ifyou have any problems visualizing what is going on here!ANSWERS: (a) 0.82 moles N2O, 0.64 moles O2, and 6.03 moles N2O4 will be present after 5.00 molesof N2O reacts.(b) No. There isn’t enough O2 for that to occur.(c) O2 will be the limiting reactant.(d) 0.39 moles of N2O and 6.46 moles of N2O4 will be in the box.(e) 23 xEXPLANATIONS:(a) How many moles of N2O, O2, and N2O4 will be present in the box IF (only) 5.00 moles (of theoriginal 5.82 moles) of N2O react according to the equation above.Answer: 0.82 moles N2O, 0.64 moles O2, and 6.03 moles N2O4 will be present after 5.00 moles ofN2O reacts.Reasoning/Explanation:First find out what changes accompany the reaction that takes place:N2O: Obviously 5.00 moles get “unformed”; that is stated in the problem.O2: From the balanced equation, the number of moles of O2 that react is always threehalves the number of moles of N2O that reacts (since the ratio is 3:2 for O2:N2O). So3/2 x 5.00 moles N2O 15/2 7.50 moles of O2 get “used up” during the reaction.N2O4: From the balanced equation, the number of moles of N2O4 formed is always equal tothe number of moles of N2O “unformed” (ratio is 2:2), so 5.00 moles of N2O4 must beformed during the reaction.So, what is present in the box at the end of the reaction is “what was there to begin with” plus(or minus) “what was formed (or used up)”:N2O: initial moles – moles used up 5.82 – 5.00 0.82 moles present at endO2: initial moles – moles used up 8.14 – 7.50 0.64 moles present at endN2O4: initial moles moles formed 1.03 5.00 6.03 moles present at end(b) Is it possible for 5.50 moles (of the 5.82 moles) of N2O to react in this scenario? Justify youranswer with calculations and/or reasoning.Answer: No. There isn’t enough O2 for that to occur.Reasoning/Explanation:To answer this, just calculate how much O2 would be required to react with 5.50 moles ofN2O, and then see if you actually have that much O2 in the box to begin with! If you haveenough O2, then it can react, if you don’t, it can’t all react!Based on the equation, we know that the number of moles of O2 needed to react with anynumber of moles of N2O is always three-halves as many (as in part (a)).

So, 3/2 x (5.50 moles N2O reacted) 8.25 moles O2 “reacted” (i.e., “needed”).How many are actually present in the box? Only 8.14 moles. So the answer is “no”. 5.50moles of N2O could not react in this scenario because the O2 would run out before 5.50 molesof N2O reacts.(c) Which reactant in this scenario is considered the limiting reactant? (Note: if you have understoodwhat you did in part (b), this question should be trivial. That is, do not start from scratch for thisproblem!Obviously, all of the O2 is going be used up before all of the N2O (because all of it is used upeven before 5.50 moles of N2O is reacted!), so O2 will be the limiting reactant.(d) If the reactants react as completely as possible according to the equation above, what will be thefinal contents of the box? (i.e., how many moles of N2O, O2, and N2O4 will be present after “thereaction is complete”). Hint: don’t forget that you started with some N2O4, so the total amount ofN2O4 will be the “amount started with” plus the “amount formed during the chemical reaction”.Answer: 0.39 moles of N2O and 6.46 moles of N2O4 will be in the box.Reasoning/Explanation:From part (b) and (c), we know that reaction will stop occurring when all of the O2 is gone(that is, we know that the oxygen is the one to “run out” first; it is the limiting reactant!) Sothat means that we KNOW how much O2 reacts .ALL of it, or 8.14 moles O2. Once youknow how much of ONE reactant (or product) IS ACTUALLY USED UP (or formed), thenyou can figure out how much of all of the other reactants and products are used up and formedas well. This is just like the question in part (a) now!What changes during the reaction:O2: 8.14 moles reacts.N2O: The balanced equation indicates that the number of moles of N2O that react isalways two-thirds of the number of moles of O2 that react, so:2/3 x 8.14 moles O2 reacted 5.4267 5.43 moles of N2O reactedN2O4: The balanced equation indicates that the number of moles of N2O4 that form isalways two-thirds of the number of moles of O2 used up (it is the same amount as themoles of N2O used up; the coefficients of N2O and N2O4 are the same!).So, 5.43 moles of N2O4 is formed.So, what is present in the box IF the reaction goes “to completion” is:O2: none (it’s all gone!)N2O: 5.82 moles initially – 5.43 moles “used up” 0.39 moles remaining.N2O4: 1.03 moles initially 5.43 moles formed 6.46 moles present at end.(e) (Assuming the reaction represented by the equation above,) if I let x represent “the number ofmoles of O2 that are used up in a certain reaction”, how would you properly represent “thenumber of moles of N2O that are used up (in that same reaction)”? In other words, how manymoles of N2O react (in algebraic terms) if x moles of O2 react (according to the equationabove)?Since the equation indicates that the number of moles of N2O that react is always equal totwo-thirds the amount of moles of O2 that react (ratio is 2 : 3 for N2O : O2), the number ofmoles of N2O that react must simply be: 23x