karatechopper
@lgbasallote
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karatechopper
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\[Fe+ H _{2}S0_{4}--> Fe _{2}(S0_{4})_{3}\]
lgbasallote
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interesting
karatechopper
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My teacher told me to always solve them by using this box......
karatechopper
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|dw:1346283514093:dw|
karatechopper
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I am sure u understand i can't do subnotes in drawing:/
karatechopper
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I am really way too stuck
lgbasallote
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okay...can you identify how many Fe you have on the left side?
karatechopper
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1
karatechopper
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which is what i put in the reactant
lgbasallote
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how many H?
karatechopper
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2
lgbasallote
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how many S?
karatechopper
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4
lgbasallote
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nope
lgbasallote
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check again
karatechopper
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oh i thought SO went together....so then S has 1
O has 4
lgbasallote
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right
|dw:1346283890519:dw|
now how many Fe, H, S and O in the products?
karatechopper
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Hold on lemme correct my box
karatechopper
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1 min
karatechopper
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Fe 2
S 1
o 4
But i got confuzzled when it has that parenthesis with a 3 out of it. So that means i multiply right?
lgbasallote
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yes. distribute that subscript into the original subscripts
lgbasallote
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i,e \[(H_2 O)_3 \implies H_6 O_3\]
karatechopper
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So....
S 3
0 12
correct?
lgbasallote
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wait....did you type this right? we're missing H in the products?
karatechopper
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Aw shoot. lemme show u i missed a part.
karatechopper
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\[Fe _{2}+H _{2}SO _{4}--> Fe _{2}(SO _{4})_{3}+ H _{2}\]
karatechopper
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There! missed out on the ending!
lgbasallote
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|dw:1346284521882:dw|
karatechopper
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No....The H stayed the same through it all....
lgbasallote
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lol oops. i jumbled it
lgbasallote
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|dw:1346284649236:dw|
there better
karatechopper
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yep
lgbasallote
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so...you have
\[\Large Fe_2 + H_2 SO_4 \rightarrow Fe_2 S_3 O_{12} + H_2\]
you'll want to balance the Os first
do you know how?
karatechopper
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no...
lgbasallote
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how many O on the right side?
karatechopper
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I know i have to multiply and change some stuff on the box, to make the reactants and products become equal
lgbasallote
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so...according to your box...how many O in the right side?
karatechopper
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12
lgbasallote
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and how many on the left?
karatechopper
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4
lgbasallote
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so what do you multiply to 4 to get 12?
karatechopper
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3
lgbasallote
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so multiply 3...
\[\large \implies Fe_2 + 3H_2 SO_4\rightarrow Fe_2 S_3 O_{12} + H_2\]
fllowing so far?
karatechopper
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Yes
karatechopper
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Now what....i get lost
lgbasallote
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so now... you solve the values again... |dw:1346285448126:dw|
do you get why?
lgbasallote
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wait....your question at first had one Fe in the reactant side
when you drew it again... it became Fe2.....so how many Fe are there really in the reactant side?
chemENGINEER
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yes very good approach....