## karatechopper 3 years ago @lgbasallote

1. karatechopper

$Fe+ H _{2}S0_{4}--> Fe _{2}(S0_{4})_{3}$

2. lgbasallote

interesting

3. karatechopper

My teacher told me to always solve them by using this box......

4. karatechopper

|dw:1346283514093:dw|

5. karatechopper

I am sure u understand i can't do subnotes in drawing:/

6. karatechopper

I am really way too stuck

7. lgbasallote

okay...can you identify how many Fe you have on the left side?

8. karatechopper

1

9. karatechopper

which is what i put in the reactant

10. lgbasallote

how many H?

11. karatechopper

2

12. lgbasallote

how many S?

13. karatechopper

4

14. lgbasallote

nope

15. lgbasallote

check again

16. karatechopper

oh i thought SO went together....so then S has 1 O has 4

17. lgbasallote

right |dw:1346283890519:dw| now how many Fe, H, S and O in the products?

18. karatechopper

Hold on lemme correct my box

19. karatechopper

1 min

20. 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?

21. lgbasallote

yes. distribute that subscript into the original subscripts

22. lgbasallote

i,e $(H_2 O)_3 \implies H_6 O_3$

23. karatechopper

So.... S 3 0 12 correct?

24. lgbasallote

wait....did you type this right? we're missing H in the products?

25. karatechopper

Aw shoot. lemme show u i missed a part.

26. karatechopper

$Fe _{2}+H _{2}SO _{4}--> Fe _{2}(SO _{4})_{3}+ H _{2}$

27. karatechopper

There! missed out on the ending!

28. lgbasallote

|dw:1346284521882:dw|

29. karatechopper

No....The H stayed the same through it all....

30. lgbasallote

lol oops. i jumbled it

31. lgbasallote

|dw:1346284649236:dw| there better

32. karatechopper

yep

33. 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?

34. karatechopper

no...

35. lgbasallote

how many O on the right side?

36. karatechopper

I know i have to multiply and change some stuff on the box, to make the reactants and products become equal

37. lgbasallote

so...according to your box...how many O in the right side?

38. karatechopper

12

39. lgbasallote

and how many on the left?

40. karatechopper

4

41. lgbasallote

so what do you multiply to 4 to get 12?

42. karatechopper

3

43. lgbasallote

so multiply 3... $\large \implies Fe_2 + 3H_2 SO_4\rightarrow Fe_2 S_3 O_{12} + H_2$ fllowing so far?

44. karatechopper

Yes

45. karatechopper

Now what....i get lost

46. lgbasallote

so now... you solve the values again... |dw:1346285448126:dw| do you get why?

47. 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?

48. chemENGINEER

yes very good approach....