Notes: Stoichiometry (text Ch. 9)

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1 Name Per. Notes: Stoichiometry (text Ch. 9) NOTE: This set of class notes is not complete. We will be filling in information in class. If you are absent, it is your res onsibility to get missing information from a fellow classmate or the chemistry website: htto J/othschem. weeblv. I. What is stoichiometry? Stoichiometry is the calculation of quantities in a chemical reaction. You must have a balanced chemical equation in order to do stoichiometry calculations. You will use a given amount of one substance in a chemical reaction to calculate an amount of another substance in that same reaction. II. Interpreting a balanced chemical equation: This equation can be interpreted as follows: 2 atoms of iron react with 3 molecules of chlorine to produce 2 formula units of iron (III) chloride. Coefficients can also be interpreted in a more useful way: MOLES! Another interpretation, based on moles: 2 Fe + 3 Cl2 2 FeCI3 2 moles of solid iron react with 3 moles of chlorine gas to produce 2 moles of solid iron (III) chloride. This is just as if we multiplied the whole equation by 6.02 x III. The Mole Ratio Concept A mole ratio can be written to compare any two substances in a reaction. Consider the example: 2 BFa + 3 H2O - I B2O3 + HP balance first! The ratio of BF3 to H2O is 2 :. In stoichiometry calculations, this ratio will take the form mol BF3 mol H2O 3 mol H2O or 2 mol BF3 Practice: Using the balanced equation above, write mole ratios for... 2 BF3 : B2O3 2. mol BF3 j mol B2O3 _ mol B2O3 or mol BFs 3 H2O :&HF 5 ol H7O (p ITVtO I H F ; B2O3 :( HF I mol B2Q3 (j) nool HF or (0 mol Hf 3 r* ol H2O (jl mol H F I mol 620 t

2 IV. Review of Mole Conversions You will need to use these conversions for many stoichiometry calculations. 1 mole = 6.02 x 1023 particles (particles can be atoms, molecules, formula units) 1 mole = X grams, where X = molar mass of the substance ex'. mo\ C0Z l. OO a CO*. * 1 mole = 22.4 L, for gases at STP WOlCilT WlCi S FrorW Review practice: 1) How many moles are in 5.0 x 1025 molecules of COa? 0XI025 x I ol COz 83 r iol G>.02 l015nnof ulcs 2) What is the mass in grams of moles of CO2? o.iso i co, x l i-ooci a co s 11.0d coz X I r$\o C0Z 1 u 3) How many moles of CO2 gas are in L at STP? loo.o COi I mol Co 2.2 4\ COi CO2. V. Basic Stoichiometry Calculations 3 steps to a basic stoich. problem: LEARN THESE STEPS! In Identify the given, and convert it to moles. Use one of the three mole conversion factors above. If Ifs already in moles, you don t need this step I Identify the unknown, and do a mole-to-mole ratio between given and unknown using the coefficients from the balanced equation. This is the key step: it gets you from moles of given to moles of unknown. NEVER skip this step! NfeoJ s-tep s ' Convert the unknown to the unit specified in the problem. If you are asked to solve for moles, you don t need this step! Q Practice with basic stoichiometry problems: AI2O3 J_AI Corrcrf vnh balance first! Mole - Mole ( 6+tp) 1. How many moles of Al could be produced from 13.0 moles of AI2O3? 3.o?l Xp3 -Jdj AI ows fta+io Zifi.O rvol Al ' TfYo> l 2. How many moles of AI2O3 would be required to produce 5 moles of Al? 5. rmx 2 01 ; i /M203 fs o Al IS-F*

3 2AI203 > 4 Al + Oi <f> r\/ / Mole - Particle (1 mole = 6.02 x 1023 molecules, atoms, formula units) 3. How many molecules of AI2O3 are needed to produce1.85 moles of Al? 2-02XI0*3moltt(/frS A 2 3 I-85 X ~ X- rvtstw l I viott-'m-g. TXio23 rvjor'cuic A Oj 4. If 1.20 x 1021 atoms of Al are made, how many moles of O2 are also produced? (0.0OI50 mol you - Mass (1 mole = X grams) How many grams of O2 would be produced from 22jmojes ofal203? 2 fv w 2s.F How many moles of Al would be produced from 890 grams of AI2O3? ( II ol / ) you h jj S Mole - Volume (1 mole = 22.4 L) L of O2 gas at STP are produced when how many moles of AI2O3 decompose? 2 mol Al Og _ 22 H 3 O.Sfo molf- 8. If the reaction used 65.7 moles of AI2O3, how many liters of O2 gas are produced at STP? ( ZilO L Oj) yo hrtj

4 Mixed practice: aj steps!!! Use the following equation 2 AI2O3 Al + 3 O2 balance first! 9. How many grams of O2 would be produced from S cjrams qf Q? ( lig AIz03 I n I AlxQ? lol. lg Al20i 3 mol Qz Sl.WgQ Z mol AI2.O3 I m l 02 30Zg 0Z 10. How many liters of O2 gas at SIR will be produced if 50.0 g of Al are also produced? (3U L you +rj 11. How many grams of AI2O3 would be needed to produce 1258 grams ofal? Al /molai 2tmo /Uz c)( l 3 AlzO 3 m l U X~l,olfllz How many grams of oxygen gas are produced when 7.50 x 1022 atoms of Al are also produced? (ZMgOz) 13. If I S qrams of AI2O3 react to produce Al, how many atoms of Al are produced? 12-5 AUO, Imo1 AI2-3 y X (pqzx,023ato-/??ll 1ZXI023I J AI2.O3 Zmo\/\\z05 ol Al More practice problems, varying # of steps: 3 CU(s) + 8 HN03(aq) 3 Cu(N03)2 (aq) + 2 NO(g) + 4 H20(l) 14. How many grams of Cu(NC>3)2 will be produced if mol Cu are available? (IHIo C (N03)2) you -fry J 15. How many liters of NO (at STR) can be produced from SO S c Cu? 5Q.( q Cu x- lmo1 V- - - LN0 v 22-1L ( O (p3. HO C / 3 vio 1 C I mol O 11. } LUO 16. How many moles of HNOa are necessary to produce 8.0 moles of H2O? you 3 (Konnol HW05)

5 VI. Limiting Reactant The Concept In a chemical reaction, one of the reactants will usually run out before the other reactant(s). At this point, the reaction will stop, because one of the reactants is used up. The reactant that is used up first is called the limiting reactant. It controls how much of the other reactants you will need, and it also controls the amount of product you can make. The other reactant(s) is/are called the excess reactant(s) because there will be some left over. Here is an example with grilled cheese sandwiches. You are given 10 pieces of cheese and 50 pieces of bread and the equation for making grilled cheese sandwiches is below: A 1 piece cheese + 2 pieces bread 1 cheese sandwich How many complete grilled cheese sandwiches can you make? lo'& x 1 I ( 5+ movn+ Of prod c+i So chcwe = L - I'd Sa aw. - 5 s-andw- 2C What is the limiting reactant (the one that ran out first)? What is the excess reactant (the one that is left over)? How many pieces of the excess reactant will be used? - Bread 20 {Start with the limiting reactant and convert to the excess reactant) I How many pieces of the excess reactant will be left over? 30 (How many did you start with and how many were used?) _ 2 O = 3 0 Icff over* I T ~ Connection to a chemical reaction: sti cftd t 2 Al(s) + 3 CuCIa (aq) 3 Cu (s) + 2 AICI3 (aq) Assume that the copper (II) chloride is the limiting reactant. Which will be present in the reaction container once the reaction has sto ped completely? A\C\3 EYcry+hir>j excep-h -fine fiom inj reah n+ Classify the above reaction as one of the 5 basic types of chemical reactions. 3)f jlc rc.ploicfrr) rrf~ 5

6 Calculations with Limiting Reactant In basic stoichiometry problems, you are provided with only one given quantity. In limiting reactant problems, you are provided with amounts of both reactants. Before you can solve the stoichiometry problem, you have to find out which of the two reactants is your limiting reactant. The limiting reactant is used as the given quantity in all calculations, because it controls the amounts of the other reactant(s) used and products made. It is possible to have the two reactants run out at the same time, and if this happens, the amounts of reactants are called stoichiometric amounts. In this special case, you can use either reactant for the limiting reactant; it does not matter. How to find the limiting reactant: 1) Start with the given amount of one of the reactants (we ll call this reactant A), and solve for the amount of one of the products (we ll call this product C). 2) Next, start with the given amount of the other reactant (reactant B) and solve for the same product (product C). 3) The reactant that produces the; least jamount of product is your limiting reactant. -A-You must then use the limiting reactant to work any additional stoichiometry problem(s). Practice: Al(s) + _CuCl2(aq) _5_ Cu (s) + AIC taq) balance first! 17. Find the limiting reactant if 0.25 mol Al and 0.35 mol CuCb are available mol Ajx 0.38noolCd Z 0.35 motgug42x -Lfl 0! Cv/ v»ol Co SmaUtri 5 C CI Limiting Reactant: & 6

7 XC nr«.c*an+- & L 2 Al(s) + 3CuCl2(aq) 3 Cu (s) + 2AICl3(aq) 18. (a) Find the limiting reactant if 6.20 g Al and 48.5 g C11CI2 are available. - olc$ 0 A+iO (?) corrtc- ni q Al x l olal y, ' v>0l v 6>5«l1( 3Cv, jn fv - i'vnalj r- a *' 7 ivioi 2 viol Al I ol I rw) Cv C * A Iniol CvCIz 3rM*/ C>/ 48.5 g CuCh x cvfl* 3 01 c /di x 315 9C,. 22. qci/ I Cu J Limiting reactant: Al (b) How many grams of copper will be produced? (start with limiting reactant from a) I o I I 3 ol Cv (* 20<j Al x X - a-cv= / Zl. gc*/ 1 20».432 rvivl Al Imol C ' (c) How many grams of the excess reactant will be used? (start with limiting reactant) Lp-20 l / viol AI 3 viol CvG z 2(. 82g I Z vioi A /3. 23 CvC/2 _ (.3q I ol CvCl How many grams are left over? 2- g Sfeirrtdi /, ifg.5 - ( 3 = 2.24 /er ove! you 2 H2 (g) + 02 (g) -> 2 H2O (g) 19. (a) Find the limiting reactant if 50.0 g each of hydrogen and oxygen gases are available. Limiting reactant: z. (b) What mass of water can be produced from this reaction? 5(0.5 cj 2O) (c) How many grams of the excess reactant are used up? (o-30cj How many grams are left over? 3 7_3 Hz) 7

8 VII. Percent Yield of a Chemical Reaction When we calculate amount of product in a stoichiometry problem, the calculated answer we get is called the theoretical yield. This is because it is the amount we would get theoretically, if our experiment went perfectly and we had no experimental errors or limitations in our procedure. However, in the real world we do not often get it perfect. The amount of product we get when we do the reaction in lab is called the experimental yield. We can compare our experimental yield with the theoretical yield to see how well we did. To do this, we use the formula below to get a percent yield, or the percentage of the theoretical yield that we obtained in lab.*the closer we get to 100%, the better we did -ft- Percent Yield = Experimental Yield x 100 Theoretical Yield Sample Problems: 20. The following reaction is performed by a student: Mg(S) + 2HCI(aq) MgCbfaq) + H2(g) She starts with 0.25 g Mg (and excess HCI) and she obtains 0.90 q of MqC product (that s her experimental yield). What is her percent yield of MgCte? (a) First, calculate the theoretical yield of MgC with stoichiometry. 0.25cj I ol 2* M<j I m l Mcjdz. Cj M CIZ_ I mol Mg I ol gcit MgCIi (b) Next, calculate the percent yield of MgCte / yitld = x 100 = Another student obtains 0.74 q of MqCte product, starting with the same 0.25 g Mg. What is his percent yield of MgC? g x , The firstsfudtnf (W20) ob-faincd higher 7. yr?ldj there F0irc his/ her exper/poer-h ha cl less r o. Ih Clo e '/.yield is fv ) - h getter! -fir 8