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karatechopper
 4 years ago
@lgbasallote
karatechopper
 4 years ago
@lgbasallote

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karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0\[Fe+ H _{2}S0_{4}> Fe _{2}(S0_{4})_{3}\]

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0My teacher told me to always solve them by using this box......

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1346283514093:dw

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0I am sure u understand i can't do subnotes in drawing:/

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0I am really way too stuck

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay...can you identify how many Fe you have on the left side?

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0which is what i put in the reactant

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0oh i thought SO went together....so then S has 1 O has 4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right dw:1346283890519:dw now how many Fe, H, S and O in the products?

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0Hold on lemme correct my box

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0Fe 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes. distribute that subscript into the original subscripts

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i,e \[(H_2 O)_3 \implies H_6 O_3\]

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0So.... S 3 0 12 correct?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait....did you type this right? we're missing H in the products?

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0Aw shoot. lemme show u i missed a part.

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0\[Fe _{2}+H _{2}SO _{4}> Fe _{2}(SO _{4})_{3}+ H _{2}\]

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0There! missed out on the ending!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1346284521882:dw

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0No....The H stayed the same through it all....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol oops. i jumbled it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1346284649236:dw there better

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so...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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how many O on the right side?

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0I know i have to multiply and change some stuff on the box, to make the reactants and products become equal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so...according to your box...how many O in the right side?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and how many on the left?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so what do you multiply to 4 to get 12?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so multiply 3... \[\large \implies Fe_2 + 3H_2 SO_4\rightarrow Fe_2 S_3 O_{12} + H_2\] fllowing so far?

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0Now what....i get lost

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so now... you solve the values again... dw:1346285448126:dw do you get why?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait....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
 4 years ago
Best ResponseYou've already chosen the best response.0yes very good approach....
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