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\[Fe+ H _{2}S0_{4}--> Fe _{2}(S0_{4})_{3}\]
interesting
My teacher told me to always solve them by using this box......

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Other answers:

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

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