!"#$%&%$#'"()*+ $ ,'-./)012.#'3"4)
53./6)5'7$").)8./."-$9)-+$,'-./)$12.#'3")3%)'":3%,.#'3"):%3,);+'-+).)-+$,'-./)$12.#'3")-.")8$)
;%'##$"<)9$4-%'8$)'#4),$."'"().#)#+$)&.%#'-2/.#$<),3/.%)."9),.-%34-3&'-)/$7$/4=)
0>.,&/$)?6!!!!!!!!!!!!!!!!!!2H
2
(g)!!!!!!!!!!!+!!!!!!!!!O
2
(g)!!!!!!!!!!!"!!!!!!!!!!2H
2
O(g)!
Particulate!Level:!!!!!!!2!molecules!!!!!!!!!!!1!molecule!!!!!!!!!!!!!!!2!molecules!
Molar!Level:!!!!!!!!!!!!!!!!2!moles!!!!!!!!!!!!!!!!!!1!mole!!!!!!!!!!!!!!!!!!!!!!2!moles!
Macroscopic!Level:!!!!!4.04!g!!!!!!!!!!!!!!!!!!!!!!!!!!32.00g!!!!!!!!!!!!!!!!!!!!!!!!!!36.04!g!
!
0>.,&/$)@6!!!!!!!!!!!!!!!!!!!!!2C
2
H
6
(g)!!!!!+!!!!!!7O
2
(g)!!!!!!!!!!"!!!!!!!4CO
2
(g)!!!!!!!!!!!!!+!!!!!6H
2
O!(l)!
Particulate!Level:!!!!!!!2!molecules!!!!!!!!!!!7molecules!!!!!!!!!!!4!molecules!!!!!!!!!!!!6!molecules!
Molar!Level:!!!!!!!!!!!!!!!!2!moles!!!!!!!!!!!!!!!!!!7!moles!!!!!!!!!!!!!!!!!4!moles!!!!!!!!!!!!!!!!!!!!6!moles!
Macroscopic!Level:!!!!!60.16!g!!!!!!!!!!!!!!!!!!224.00!g!!!!!!!!!!!!!!!176.04!g!!!!!!!!!!!!!!!!!!!108.12!g!
!
A2."#'#B)C$/.#'3"4+'&4)'")*+$,'-./)C$.-#'3"4<)DE#3'-+'3 ,$#%BF)
53./6)5'7$").)-+$,'-./)$12.#'3"<)3%).)%$.-#'3"):3%);+'-+)#+$)$12.#'3")'4)G"3;"<)."9)#+$)"2,8$%)3:)
,3/$4)3:)3"$)4&$-'$4)'")#+$)%$.-#'3"<)-./-2/.#$)#+$)"2,8$%)3:),3/$4)3:)."B)3#+$%)4&$-'$4=)
*3"4'9$%)#+$)$12.#'3"6)
4NH
3
(g)!+!5O
2
(g))))"!!4NO(g)!+!6H
2
O(g)!
1. )H3;),."B)I
@
),3/$-2/$4).%$)%$12'%$9)#3)%$.-#);'#+)JKL),3/$-2/$4)3:)MH
J
N)
O"4;$%6))On!the!particulate!level,!
4!molecules!of!NH
3
!react!with!5!m o lecu les!o f!O
2
!
That!is!!!!!!!!
#!$%&'()&'*!!+
,
-!$%&'()&'*!./
0
!!!!OR!!!
-!$%&'()&'*!!!./
0
#!$%&'()&'*!+
,
!
!
Recall!that:!Required!number!and!units!=!Given!number!and!units!x!conversion!factor!
Therefore,!
#!O
2
!molecules!=!308!NH
3
!molecules!!x!!
1!234567458!!+
,
9!234567458!./
0
))P)JLQ)I
@
),3/$-2/$4)
2. If!3.20!m o le s !o f!N H
3
!react!accord ing !to!the!a bo ve!e qu atio n,!ho w !m an y!m ole s!of!H
2
O!will!be!
produced?!
O"4;$%6))The!unit!path!is!mol!NH
3
!"! mol!H
2
O!
Therefore,!use)!!
:!$%& !/
,
+
-!$%& !. /
0
!!!!as!the!conversion!factor!
Hence,!mole!H
2
O!=!3.20!mol!NH
3
!!x!!
:!$%&!/
,
+
-!$%& !. /
0
!!!!=!!4.80!mol!H
2
O!
!
3. How!!many!moles!of!oxygen!are!required!to!burn!2.40!moles!of!ethane,!C
2
H
6
?!
!
Answer:!First,!write!a!balanced!equation!for!the!reaction!described.!
H'"#6))C$,$,8$%)#+.#)-3,824#'3")3:)+B9%3-.%83")./;.B4):3%,4)*I
@
)."9)H
@
I)
!
2C
2
H
6
(g)!!+!!7O
2
(g)!!"!!4CO
2
(g)!!+!!6H
2
O!(l)!!!!
!
The!conversion!factor!from!the!relationship!between!the!given!and!the!wanted!quantity!using!
the!coefficients!in!the !equ atio n!is!!
;!$%& !+
,
<!$%& !=
,
/
>
!
!
Hence,!mol!O
2
!=!2.40!mol!C
2
H
6
!!x!!!
;!$%& !+
,
<!$%&!=
,
/
>
!=!8.40!mol!O
2!
!
!
R.44)*./-2/.#'3"4)
!!Mass!calculations!may!involve:!
a. !Writing!!a!balanced!equation!
b. Calculating!molar!masses!from!chemical!formulas!
c. Using!molar!masses!to!change!from!mass!to!moles!or!from!moles!to!m as s!
d. Using!the!equation!to!change!from!moles!of!one!species!to!moles!of!another!
The!mass!–to-!mass!path!is:!
Mass!of!given!"!moles!of!given!"!moles!of!wanted!"!mass!of!wanted!
Mass!given!x!!!
$%&!?@A'B !
$C**!?@A'B
!x!!!
$%&!DCBE'F
$%&!?@A'B
!!!x!
$C**!DCBE'F
$%&!DCBE'F
!
0>.,&/$6))*./-2/.#$)#+$)"2,8$%)3:)(%.,4)3:)3>B($")#+.#).%$)%$12'%$9)#3)82%")?QQ)()3:)$#+."$)
'")#+$)%$ . - #'3 " 6)2C
2
H
6
(g)!!+!!7O
2
(g)!!"!!4CO
2
(g)!!+!!6H
2
O!(l)!!!!
O"4;$%6)))
M2,8$%)3:)(%.,)I
@
)P)?QQ)()C
2
H
6
!x!!!!
G!$%& !=
,!
/
>
HIJI;!?!=
,!
/
>
!x!!!!
;!$%& !+
,
<!$%& !=
,!
/
>
!!!x!!
H<JII!?!+
,
!
G!$%& !+
,
!!=!!577!g!O
2!
)
0>.,&/$6))H3;),."B)(%.,4)3:)3>B($").%$)%$12'%$9)#3)82%")J=QK),3/$4)3:)/'12'9)+$&#."$<)
*
S
H
?T
D/FN)
O"4;$%6!First,!write!a!balanced!equation.!
C
7
H
16
(l)!!+!11O
2
(g)!!!"!7CO
2
(g)!+!8H
2
O(l)!
g!O
2
!!=!!3.50!mol!C
7
H
16
!!x!!
GG!$%&!+
,
G!$%& !!=
K
/
L>
!!x!!
H<JII!?!+
,
G!$%& !+
,
!!!=!!!1,230!g!O
2
!
!
0>.,&/$6))H3;),."B)(%.,4)3:)*I
@
);'//)8$)&%392-$9)8B)82%"'"()TT=K)()*
S
H
?T
)).--3%9'"()#3)#+$)
$12.#'3"6)C
7
H
16
(l)!!+!11O
2
(g)!!!"!7CO
2
(g)!+!8H
2
O(l)?!
O"4;$%6)
g!CO
2
!!=!66.0!g!C
7
H
16
!!x!
G!$%& !!=
K
/
L>
!
GIIJ<I!?!=
K
/
L>
!!x!!
;!$%& !=+
,
G!$%& !!=
K
/
L>
!!x!!
--JIG!?!!=+
,
G!$%& !!=+
,
!!!=!203!g!CO
2
!
!
! !
U','#'"()."9)0>-$44)C$.-#."#4<)V+$3%$#'-./)W'$/9<)O-#2./)W'$/9)."9)X$%-$"#)W'$/9)
Y$:'"'#'3"46)
Limiting!reactant:!the!reactant!that!is!comp le te ly !us e d !up !d u rin g !th e !re ac tio n !is!c a lled !t he !limiting!
reactant.!
Excess!reactant:!the!reactant!that!some!of!it!remains!unreacted!after!the!reaction!is!complete.!
Theoretical!Yield:!the!amount!of!product!formed!from!the!complete!conversion!of!the!given!amo un t!of!
reactant!to!prod u ct.!!The ore tical!yield !is!alwa ys!a!-./-2/.#$9)12."#'#B!based!on!the!stoichiometry,!(ratio)!
of!the!reaction!equation.!
Actual!Yield:!a!measured!quantity!determined!by!experiment.!Factors!such!as!impure!reactants,!
incomp le te !re a ct ion !a n d !sid e !re ac tio n s !ca u se !th e !a ctu a l!y ield !to !b e !le ss !th a n !th e!t h eo re tic a l!yie ld .!
Percent!Yield:!this!is!the!actual!yield!expressed!as!a!percentage!of!the!theoretical!yield.!
M!Yield!!=!!!
C(E)C&!N@'&F
EO'%P'E@(C& ! N@'&F
!!x!100!
Z"9$%4#."9'"()U','#'"()%$.-#."#!
a. !How!many!pairs!of!gloves!can!you!assemble!from!20!left!gloves!and!30!right!gloves?!
O"4;$%6!!20!pairs!Q!th eo retica l!yield!
!
b. How!many!unmatched!gloves!will!be!left!over?!
O"4;$%6!10!right!gloves!Q!excess!amount!
!
c. Which!hand!will!these!unmatched!gloves!fit?!
O"4;$%6!Right!Q!excess!reactant!
!
d. What!prevented!you!from!assembling!more!than!20!pairs?!
O"4;$%6!Left!gloves!Q!limiting!rea ct an t !
!
H3;)V3)E3/7$)U','#'"()C$.-#."#)X%38/$,4!
There!are!two!methods:!
a. Smaller-Amount!method!
b. Comparison-of-Moles!Method!
We!shall!use!the!smaller-amount!method!
X%3-$92%$6!
1. !Calculate!the!amount!of!product!that!can!be!formed!by!the!initial!amount!of!each!reactant!
a. The!reactant!that!yields!the!smaller!amount!of!product!is!the!/','#'"( )%$.-#."# = !
b. The!smaller!amount!of!product!is!the!amount!(#+$ 3%$ #'-./) B'$/9)!that!w ill!be!formed !when !
all!of!the!limiting!reactant!is!used!up.!
2. Calculate! the! amount! of! excess! reactant! that! is! used! by! the! total! amount! of! the! limiting!
reactant.!
3. Subtract!from! the!am ount! of! excess! reactant!present! initially,!the! amount! that! is!used! by! the!
limiting!re a ct an t .!!Th e !d iffe re n ce !is!t h e!a mount!o f!e xc e ss !rea c ta n t!th a t!is !le ft.!
0>.,&/$6))*./-2/.#$) #+$) ,.44) 3:) ."#',3"B) D!!!F) -+/3%'9$<) E8*/
J
<) #+.#) -.") 8$ ) &%392- $9 ) 8B) #+$ )
%$.-#'3")3:)?@[)()."#',3"B<)E8<)."9)?KT)()-+/3%'"$=))H3;),2-+)3:)#+$)$>-$44)%$.-#."#)'4)/$:#N!
O"4;$%6)
) ) ) 2Sb!!!+3Cl
2
"2SbCl
3
!
Molar!mass!(g/mol)!!!!121.8!!!!!!!!!71!!!!!!!!!!!!!228.3!
Mass!Used!(g)!!!!!!!!!!!!!129!!!!!!!!!!106!
Moles!Used,!(mol)!!!!!
G<GJR!?
G<S!?T $%&
GI:!?
;G!?T $%&
!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1.059!mol!!!!!!!1.492!mol!
a.!From!balanced!equation,!
!!!!2!mol!Sb!"!2!mol!SbCl
3
!
!!!!1.059!mol!Sb!"!xmol!SbCl
3
!
!!!Hence,!x!mol!SbCl
3
!!!=!1.059!molSb!
!!!b.!!!3!mol!Cl
2
"2mol!SbCl
3
!
!!!!!!!!!1.492!mol!Cl
2
"!x!mol!SbCl
3
!
Hence,!x!mol!SbCl
3
!=!!
GJ-S<!$%& !=U
,
!V!<!WXU!YZ=U
0
H!$%& !=U
,
=!0.995!mol!SbCl
3
!
Comparing!amounts!of!products!from!each!reactantshows!that!Cl
2
!produces!less!SbCl
3
!than!Sb.!
Therefore:!
(i) Cl
2
!is!the!lim it in g!r ea ct an t .!
(ii) Sb!is!the!excess!reactant.!
(iii) 0.995!mol!SbCl
3
!produced!from!the!limiting!reactant!is!the!#+$3%$#'-. /)B'$/9.!
!
0.995!mol!SbCl
3
!x!=!!
<<RJH!?!YZ=U
0
G$%&! YZ =U
0
!!=!227!g!SbCl
3!
!
!!!c.!Excess!amount!of!Sb!
!!!!!!To!find!out!how!much!of!Sb!is!left,!first!find!how!much!Sb!was!used!to!react!with!all!106!g!Cl
2
!
!!!!!!Amount!Sb!Used:!
!!!!!!106!g!Cl
2
!x!
G!$%& !=U
,
;G!?!=U
,
!!x!!
<!$%& !YZ
H!$%& !=U
,
!!!x!!!
G<GJR!?!YZ
G!$%& !YZ
!!=!121!g!Sb!
!!!!!Amount!left!(excess)!=!129!g!(initial)!–!121!g!(used)!!=8!g!Sb!
X$%-$"#)W'$/9!
A!student!carried!out!the!reaction!in!the!last!example!and!obtained!210!g!SbCl
3
.!!Wh a t!is!the!stu de n t’s!
percent!yield?!
O"4;$%6)Actual!yield!=!210!g!SbCl
3
!
!!!!!!!!!!!!!!!!Theoretical!yield!=!227!g!Sb Cl
3
!
M!Yield!=!
C(E)C&!N@'&F
EO'%P'E@(C& ! N@'&F
!!!!x!!!100!
MYield!!=!
<GI!?!YZ= U
0
<<;!?!YZ= U
0
!!!x!!!!100!!!!!=!!93%!
!
!
!
!
!
!
!
!
!
)
!
)