stoikiometri
DESCRIPTION
gfxgbxgTRANSCRIPT
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6
!$,,,
2 16 18 Hidrogen Oksigen Air
g g g+
-
,, -..
6 12 10
100 900cHidrogen Oksigen Air H x J
g g
+ =
( )2
26 8
610
16
10
12 10 3 10
12 10 1,33 109 101,33 10 (kecil sekali)
E mc
x m x
xm x kg
x
m x g
=
=
= =
=
-
+ ,
2+,%31 3%,45
6
Air mengandung hidrogen 11,19%1:8
oksigen 88,81%
-
, +
2,17%,1%117 3%, 1 45
6
-
&.%8'
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8)%)
:)%&&
:)%'
9:%'
):%-
-
+
2; 61166%, , 66, ,45
-
6
+ 1 '8608--%--&9
8 : 1
89,88 11,11
99,99
14,5742,86
75 75
14,57 86,42
25 75
metana
gg
g
xg
gg
gg
airoksigenhidrogen
monoksidakarbonoksigenkarbon
hidrogenkarbon
+
+
+
-
+
!)
-
+
$ 1 ( %1&%..( )8%84
-
2 2 2
2
2 2
2
2
1
2
biloks 0 0 2 -21 1 elektron1 1 ekivalen 1 ekiv 11 21
2 2
H O H O
mol H molmol H massa H grmol H gram
mol O ekivale
+
+
=
=
2
2
1 ekiv 81
162
n
massa O grmol O gr
=
=
-
+1 =
2+ 1%, $1 ,
, 1 45
6
volvolvolamonianitrogenhidrogen
volvolvolairuapOksigenHidrogen
2 1 3
2 1 2
+
+
-
$
2+ 1%$1145
6
molekulmolekulmolekulvolvolvol
airoksigenhidrogen
2 1 2 2 1 2
+
-
:1 1 misal molekul bola=
( )ada teori yang dilanggarmisal : 1 molekul = 2 bola
( )tidak ada teori yang dilanggar
2
-
!
&4 ,1,4
04 4
)4 ;17%, , 14
,$
( )hRUapxRapatrelatifmolekulmassa 2 =
-
"*
2&-:&, ,6&0, 45
>%
12 121
=
Catomsatumassa
unsuratomsatumassaAr
valensiekivalenmassaAr x =
-
+
&4 !,"&(&-*
6
,1.%09?
$ )(%)4 1,@
gram.Kkal
4,6 x
gram.KJ
8,26 x
jeniskalorkirakiraAratau
jeniskalorkirakiraAr
-
;7
04 633"&(8(*
2/A1,17A"/B+*
27 kira-kira 112
0,24112 3 (bilangan bulat)38,3
Ar yang tepat = 3 x 38,3=114,9
Ar
valensi
=
-
6
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F1, 60 4069 71 69.8100 100
x x
= + =
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G C9.
G C9.?
H1%&G C9.
( )( )
massa gmol
gmassa molar
mol=
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61&&4.-,784:#,+4 @
;7 5.65.6 1 0.2522.4
11.090.25 ( / )11.09
44 /0.25
LL x mol molL
gmol
massa molar g mol
massa molar g mol
=
=
= =
-
" !!
6 10948 09 0#@
;7
mol mmolmolar
liter mL= =
24.5 0.2598 /0.25 0.125
2M
gmol mol
g molmolC ML
= =
= =
-
!!
6
EGG09 "GC0)IC)0IC&:*
;7
x % x 100%jumlah atom ArunsurMr
=
2 23% 100% 32.4%142xNa x= =
-
17
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1
04 ,
,
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60'0 J 0) 6
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60'0 J 0) 6
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8 JJ60'0J) 0)06
)JJ) 9J9 0
60'0 J) 0) 06
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60'0 J) 0)J 0 06
)J J) 9 J8
11H = 3H =
6H = 8H =
-
$
$
$1, J
$1, , J
6
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6 J J 6
& &
& J C&$
& 6 C&$
& 6 C):48
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$31,
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=08 0=
J&. : J9
=08 C&(0
&=08 C&(0
&$=08 C"&(0?:*
massa 1 mol oksidator3. massa ekiv oksidator = jumlah mol elektron yang diterima
-
B B
. 0 J0
BC8:
&BC8:
&$BC"8:?0*
massa 1 mol reduktor4. massa ekiv reduktor = jumlah mol elektro yg dilepaskan
-
+,
+1,,
60M J 0 0M
0(4: '499
:849 &:
.49)( .40)0
!
.40&- .40)0
-
%3/,,,14!M4
;%M CM
C.49)(
M C.49)(":849J&:*
C)84:
-
+
6 60 9 J )0 060J0 0
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0( )0
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massa produk nyata% hasil = 100%massa produk menurut perhitungan
x
-
;%,,%0%,
60 C
,60 C:4-:
%
2 g0.37 mol x 44 10.85 gram3 mol
x
=
6.96% hasil = 100% 64%10.85
x =