ࡱ> XZWt#` ΐbjbj !5GGGGT\Go2HIIIIOLOLOL4o6o6o6o6o6o6o$ qhss@Zo9OLK^OLOLOLZoII9oiNiNiNOLI(I4oiNOL4oiNiN,gv6iIH Qu6GLgio0ohsMRs4iisihOLOLiNOLOLOLOLOLZoZoMNOLOLOLoOLOLOLOL)// Limiting Reagent and Theoretical Yield Determining Limiting Reagent and Theoretical Yield using ICE Tables Here is a general procedure to follow for limiting reagents: Check to be sure you have a balanced equation and set up the ICE table. Example 4 NH3(g)+ 5 O2(g)(4 NO(g) +6 H2O(g)ICE Begin filling in the ICE table by writing in the initial mole amount of each species, which can be converted from grams using molar mass. (Fill in the I linesthe first line under the chemical equation). Example How many grams of NO form when 1.50 g of NH3 reacts with 1.85 g of O2?  EMBED Equation.3  4 NH3(g)+ 5 O2(g)(4 NO(g) +6 H2O(g)I0.0881 mol 0.0578 mol 0 mol0 mol CE Fill in line C of the ICE table, the change in the number of moles of all species. This is represented by the variable x. Remember to bring your coefficient in the balanced equation into the C row and multiply it by x. The change for reactants will by minus, since you are losing some of the reactant, and the change for products will be plus, since you are forming product. Example 4 NH3(g)+ 5 O2(g)(4 NO(g) +6 H2O(g)I0.0881 mol0.0578 mol0 mol0 molC-4x-5x+4x+6xE Determine how many moles will it require to use up each of the reactants and choose the smallest value to use for x Example 0.0881 mol NH3 (g) -4x = 0 ! x = 0.0220 mol 0.0578 mol O2 (g) -5x = 0 ! x = 0.0116 mol (gives smaller x therefore limiting!) Fill in line E of the ICE table, this is by adding the initial moles to the change in moles from rows one and two of your table Example 4 NH3(g)+ 5 O2(g)(4 NO(g) +6 H2O(g)I0.0881 mol0.0578 mol0 mol0 molC-4x-5x+4x+6x0.0881 mol-4(0.0116 mol)0.0578 mol-5(0.0116 mol)4(0.0116 mol)6(0.0116 mol)E0.0418 mol 0 mol 0.0463 mol 0.0694 mol The extra line in the ICE Table just demonstrates my thinking process Use the information from row E to determine other information such as the mass of product produced or the amount of excess reactant Example How many grams of NO form when 1.50 g of NH3 reacts with 1.85 g of O2?  EMBED Equation.3  Example How much of the excess reactant remains after the limiting reactant is completely consumed?  EMBED Equation.3  Questions (Please try these problems using the ICE table Approach) A reaction mixture contains 70.0 g of Fe3O4 and 12.0 g of O2 and the reaction below is allowed to occur. 4 Fe3O4 (s) + O2 (g) ( 6 Fe2O3 (s) How many grams of Fe2O3 will be generated? How much of each reagent will remain after the reaction has gone to completion? If 59.6 grams of Fe2O3 are produced, what is the percent yield of the reaction? Sulfur dichloride reacts with sodium fluoride according to the equation: 3 SCl2(l) + 4 NaF(s) ( SF4(g) + S2Cl2(l) + 4 NaCl(s) If 30.00 g of SCl2 and 20.00 g of NaF react to yield 8.00 g of SF4, what is the percent yield of SF4? Solutions A reaction mixture contains 70.0 g of Fe3O4 and 12.0 g of O2 and the reaction below is allowed to occur. 4 Fe3O4 + O2 ( 6 Fe2O3 I0.302 mol 0.375 mol 0 mol C-4x-x+6xE0 mol 0.299 mol0.453 mol Initial moles of reactants to insert above  EMBED Equation.3  Possible Xs 0.302 mol = 4x x = 0.375 mol X =0.0755 mol choose smaller value How many grams of Fe2O3 will be generated?  EMBED Equation.3  EMBED Equation.3  How much of each reagent will remain after the reaction has gone to completion?  EMBED Equation.3  If 59.6 grams of Fe2O3 are produced, what is the percent yield of the reaction?  EMBED Equation.3  Sulfur dichloride reacts with sodium fluoride according to the equation: 3 SCl2(l) + 4 NaF(s) ( SF4(g) + S2Cl2(l) + 4 NaCl(s) If 30.00 g of SCl2 and 20.00 g of NaF react to yield 8.00 g of SF4, what is the percent yield of SF4?  3 SCl2(l) + 4 NaF(s) ------> SF4(g) + S2Cl2(l) + 4 NaCl(s) I0.2915 mol0.4763 mol------------(-3x-4x+x+x+4xE00.0876 mol0.09717 mol'DGk     " # ( ) 0 1 8 9 @ A B    E F ^ _ d e x ƻѻƍ~vlvlv_vjhShS5UhShS5H*hShS5 hSh6EhSh6E5hShx`5 hShS% jhShSCJaJmHnHuhShSCJH*aJhShSCJaJhSh6ECJaJhShx`CJaJhSh5\hSh#CJaJhShCJaJh%'k    ( $$If[$\$a$gdSh[$\$^hgdx` & F[$\$gdS@&gd$a$gd$a$gd ͐( ) + 5"$$If[$\$a$gdSkd$$Iflֈ`, !\( t0644 laytS+ , - . / 0 $If[$\$gdx`0 1 3 5"$$If[$\$a$gdSkdw$$Iflֈ`, !\( t0644 laytS3 4 5 6 7 8 $If[$\$gdx`8 9 ; 5"$$If[$\$a$gdSkd$$Iflֈ`, !\( t0644 laytS; < = > ? @ $If[$\$gdx`@ A B  5% & Fgd6Eh[$\$^hgdx`kde$$Iflֈ`, !\( t0644 laytS    a b | } ~ $$If[$\$a$gdcigdSgd6E x y z { } u | } 6 a ǼǼǼǼǕǼǼǼǀxs h"5hShS5hShsOJQJ^Jh6ECJaJhSh6E5% jhShSCJaJmHnHuhShSCJH*aJhShSCJaJ hShSjhShS5UjhShS5EHU)j?@ hShS5UVmHnHu* 5"$$If[$\$a$gdcikd+$$Iflֈ`, !\( t0644 laytci $If[$\$gdci 5"$$If[$\$a$gdcikd$$Iflֈ`, !\( t0644 laytci $If[$\$gdci 5"$$If[$\$a$gdcikd$$Iflֈ`, !\( t0644 laytci $If[$\$gdci H 50( & Fgd6Egd6Ekd$$Iflֈ`, !\( t0644 laytciH I u } ~  $$If[$\$a$gdS$$If[$\$a$gdcih[$\$^hgd6E8^8gd6E & F gd6Eh^hgd6E 5"$$If[$\$a$gdcikd$$Iflֈ`, !\( t0644 laytci $$If[$\$a$gdS 5"$$If[$\$a$gdcikd~$$Iflֈ`, !\( t0644 laytci $$If[$\$a$gdS 5"$$If[$\$a$gdcikd$$Iflֈ`, !\( t0644 laytci $$If[$\$a$gdS b 5)! & FgdS $h^ha$gdx`kdl$$Iflֈ`, !\( t0644 laytcia b c j k l m x z { } $(*BD\^`bpx~ùⴢÔùˢˌˌˈ{vhhYCJOJQJ^JaJ hs5h"hY5 hY5h"hSh"5hsCJOJQJ^JaJ#h"h"5CJ(OJQJ^JaJ( hci5h"h"5H*h"h"5 h"5#h"hs5CJOJQJ^JaJh"hS5hShsOJQJ^JhShs5'b c k l &(  $$If[$\$a$gdcih^hgdY & Fgd"gd"h^hgd"h^hgdS $h^ha$gds   EFYZ[hst456TĹϋ{shYh 5h hs5h h 5hsCJOJQJ^JaJhSh"5\h h CJaJh"h CJaJh h"CJaJh"h"CJaJ hSh"% jhSh"CJaJmHnHuhSh"CJH*aJhSh"CJaJ, "-5""$$If[$\$a$gdcikd$$Iflֈ1& !\(   t0644 layt -89?E$$If[$\$a$gdciEFHL5""$$If[$\$a$gdcikd $$Iflֈ1& !\(   t0644 layt LPQUY$$If[$\$a$gdciYZ[t5""$$If[$\$a$gdcikd% $$Iflֈ1& !\(   t0644 layt t$$If[$\$a$gdci5""$$If[$\$a$gdcikd $$Iflֈ1& !\(   t0644 layt $$If[$\$a$gdci55)! & F gd  $h^ha$gdskdg $$Iflֈ1& !\(   t0644 layt 56  $%-.$a$gd#gd#h^hgd#h^hgdY & Fgd  $h^ha$gdsT     !"#%,.0zrrlblQ jC@ h#UVmHnHujh#CJ U h#CJ h#h#5h"h#5j hYCJ EHU j@@ hYUVmHnHu hYCJ jhYCJ UhShY5H*hShY5hYCJOJQJ^JaJ hY5h"hY5hsCJOJQJ^JaJhYhs5 h#5hYhY5./_`defguv}~()*+di45VW[Ǽ jh~hs h~h~h~hsH* h~hsh#hsCJ h#hsh#hsCJaJhYCJOJQJ^JaJhsCJOJQJ^JaJ h#CJ jh#CJ Ujh#CJ EHU7\`efghi8^8gds & F 8^gds $h^ha$gd~ & Fgdsh^hgdsijYZ[\]^_`abcdefghijk h1$^hgds $1$^a$gd~1$gds & F1$gd#h^hgds[bcdfghlw  DGhinѿɸѸѥїуу h~hzj'h~h\EHUjG3QB h~h\UVjh~h\Uh~hiZ>* h~hiZ jh~h\h~h\H* h~h\ h~hsh#h~CJh#h#CJ hsCJh#hsCJ0kvw +,9$$ &#$/1$Ifa$gdiZ@ ^@ gdiZ & F )`)gd#h^hgds 9:<@Q:#$$ &#$/1$Ifa$gd#$$ &#$/1$Ifa$gdiZkd$$IflrVB RT04 laytiZ@CDHI:kds$$IflrVB RT04 laytiZ$$ &#$/1$Ifa$gd#IKR\]gh:kdM$$IflrVB RT04 laytiZ$$ &#$/1$Ifa$gdiZhijklmn$%PQ< p^gd\^gd\gd\ & Fgd\h^hgd\ #%9:;<QRefgi|}~=>QRSTWٸٝقugZjh~h\EHUj;QB h~h\UVjh~h\EHUj*󾱴5>*#<=UVWXYZ[\]^_`OQR h1$^hgdiZ 1$^gdiZ1$gdiZ & F )1$`)gd#^gd\^gd\gd\W*+LMOPXY\il}~DTbcghi֎hiZjhiZU h~h@ jh~hp,UmHnHuh~hiZ5>* h~hp,U jDh~hiZh~hiZEHh~hiZEHH* jh~h#UmHnHu jh~hiZh~hiZH* h~hiZ3 $$1$Ifa$gd+1$gdiZ+ $$1$Ifa$gd+kd!$$Ifl֞ , t"9TT04 lazytiZ $$1$Ifa$gd++ $$1$Ifa$gd+kd"$$Ifl֞ , t"9TT04 lazytiZ  $$1$Ifa$gd+0.09717 mol0.3887 mol 0.2915 mol =3x 0.4763 mol = 4x x = 0.09717 mol x = 0.1191 mol  Choose smaller value   EMBED Equation.3   EMBED Equation.3  AC+%%%1$gdiZkd#$$Ifl֞ , t"9TT04 lazytiZChjkʐː̐͐ΐgdiZ1$gdiZ ƐǐȐɐ̐͐ΐȾٺ h~h@ h+j (hiZEHU j}< hiZUVmHnHuhiZjhiZUj$hiZEHU j]}< hiZUVmHnHu 21h:p1Z/ =!"#$% `!dDRC(%@ S:psxT1o@~w$R'4%E.r+9i Ёd肔T(i*Z Eag!% :Db^H{wGW $hiBxJ 'Ϯ4n%ôk q0k:dQ[rIo&O*Cm[,\>C>Mp0S^ v1?`>>/-Z)ڹ,lۡ <03p՟ Ź/Df6pW4|?!yae-/Kx%!9c>#73w̹ nX|>.󶴦=+⠅_f/X@XVEB;RgG $v5%uM^ij1 z '~dUzV~@~]gg1=k ;<|F{۱]۱R>Siz"~_43✓\En=cwUUWQVڅjZ]լFrTYnҬ8KHnDTl}'QkoJY=lxYdG AwA>˘5ܦÒ_Ӟΰ+:41L¹[ӯ(װ|ݳհkFW7y>Q(=;q]g~׳ҟ&v'/ w¸P^-Ekhq|R4u$$If!vh555555#v:Vl t65ytSu$$If!vh555555#v:Vl t65ytSu$$If!vh555555#v:Vl t65ytSu$$If!vh555555#v:Vl t65ytSODd xB  S A? 26Bg4H2B `!6Bg4H2B@>* [xTo@H6NRӪe:TNJ%DJQA)Ul Ѳ00A21.H΂T@ Ą wg[w0ν{w{;Jh=K]l  INh.CкcpV^WDߎᙢo~Sމ:^s߻ul:2% i͛Hnx`E.#؝Db5KAT`!/3Ch=>b5KAT!`R4PxSO@~لƱK"6"E]QE,@$!01wa:2ҡ  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNPQRSTUVuY\]^_`abcedgfhijklmnopqrsvwxyz{|}~Root Entry7 Fu6[@Data O+WordDocument6ObjectPool9Qu6u6_1075724184FQu6Qu6Ole CompObjfObjInfo  #$%&'()*-256789<?@ABEHIJKLORSTUVWX[^_`abcdefhijklmopqrt FMicrosoft Equation 3.0 DS Equation Equation.39qcnIzI 1.50 g NH 3 1 mol NH 3 17.03g NH 3 =0.0881 mol NHEquation Native _10757245261 FQu6Qu6Ole  CompObj f 3 1.85 g O 2 1 mol O 2 32.00g O 2 =0.0578 mol O 2 FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo Equation Native $_1075725072FQu6Qu6Ole cnIzI mass of NO formed = 0.0463 mol NO30.0 g NO1 mol NO=1.39 g NO FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjfObjInfoEquation Native 4_1112617799 FQu6Qu6cnIzI 0.0418 mol NH 3 17.03 g NH 3 1 mol NH 3 =0.7106 g NH 3  remain FMicrosoft Equation 3.0 DS EqOle CompObjfObjInfo!Equation Native "$uation Equation.39qmIyI 70.0 g Fe 3 O 4 1mol Fe 3 O 4 231.6 g Fe 3 O 4 =0.302 mol Fe 3 O 4 12.0 g O 2 1mol O 2 32.00 g O 2 =0.375 mol O 2 FMicrosoft Equation 3.0 DS Equation Equation.39qmIyI _1112620035'FQu6Qu6Ole +CompObj,fObjInfo.Equation Native /$_1112620089"FQu6Qu6Ole 0CompObj 1f FMicrosoft Equation 3.0 DS Equation Equation.39q8IԟI 0.453 mol Fe 2 O 3 159.7 g Fe 2 O 3 1 mol Fe 2 O ObjInfo!3Equation Native 4T_1112620236$FQu6Qu6Ole :3 =72.3 g Fe 2 O 3 FMicrosoft Equation 3.0 DS Equation Equation.39qpIرI 0.299 mol O 2 32.00 CompObj#%;fObjInfo&=Equation Native >_1112619984)FQu6Qu6g O 2 1 mol O 2 =9.57g O 2  remainI FMicrosoft Equation 3.0 DS Equation Equation.39q8Uai?ыTW.d؛?fǫf'~5?[u2FF;8k R<4z1x]/zlpw翅]ih)=xV/c+1o궿Y-W\YΒ-2o_P<v0OZ{7beo//}j<[ğ7Fxg!9X}ԛP1U\"$Y52x$$If!vh555T55#v#v#vT#v#v:V l0555T554ytiZ$$If!vh555T55#v#v#vT#v#v:V l0555T554ytiZ$$If!vh555T55#v#v#vT#v#v:V l0555T554ytiZDd xJ  C A? 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