karatechopper
  • karatechopper
@lgbasallote
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
karatechopper
  • karatechopper
\[Fe+ H _{2}S0_{4}--> Fe _{2}(S0_{4})_{3}\]
lgbasallote
  • lgbasallote
interesting
karatechopper
  • karatechopper
My teacher told me to always solve them by using this box......

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

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