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Jonask Group Title

Calculate the mass of iron metal (in grams) that can be prepared from 150 grams of aluminum and 250 grams of iron(III) oxide.

  • one year ago
  • one year ago

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  1. Jonask Group Title
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    \[ {F}e_2O_3+2Al\implies Al_2O_3+2Fe\]

    • one year ago
  2. Jonask Group Title
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    2Al is the limiting reagent right

    • one year ago
  3. Outkast3r09 Group Title
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    did you check to see?

    • one year ago
  4. Jonask Group Title
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    novice in chem,well i just looked at the atom it has 2 moles and Fe2O3 has 1 so yeah

    • one year ago
  5. Outkast3r09 Group Title
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    to figure out which is limiting you want to to take Grams of reactant ->mols of reactant->moles of product with both... which ever can make the least amount of moles of product is your limiting reactant

    • one year ago
  6. Outkast3r09 Group Title
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    or limiting aent

    • one year ago
  7. Jonask Group Title
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    Fe2O3 M=55 m=250 n=4.5 Al M=26 m=150 n=5.7

    • one year ago
  8. Outkast3r09 Group Title
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    alright and then there is 2 aluminum molecules per 2 iron

    • one year ago
  9. Outkast3r09 Group Title
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    so you cna make 9.16 moles of iron with 9.16 moles of aluminum

    • one year ago
  10. Jonask Group Title
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    wait 9.6 was error i took wrong mass Al n=5.7

    • one year ago
  11. Outkast3r09 Group Title
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    o wait nvm 5.7 molecules of aluminum can make 5.7 molecules for iron

    • one year ago
  12. Jonask Group Title
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    Al:Fe 1:1

    • one year ago
  13. Outkast3r09 Group Title
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    however per each iron oxide molecule you can make 2 iron oxide molecules

    • one year ago
  14. Jonask Group Title
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    yes

    • one year ago
  15. Outkast3r09 Group Title
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    so multiply 4.5*2 =9 moles of iron.... since you can only make 5.7 moles with the aluminum , Al is your limiting reagent

    • one year ago
  16. Outkast3r09 Group Title
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    now just change 5.7 moles of Fe to grams

    • one year ago
  17. Jonask Group Title
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    m=5.7*55=322g

    • one year ago
  18. Outkast3r09 Group Title
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    so that would be your answer

    • one year ago
  19. Jonask Group Title
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    so cant we say there 2Al n=5.7 but 2 moles n=2(5.7)

    • one year ago
  20. Jonask Group Title
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    answer not correct i think i made a mistake

    • one year ago
  21. Jonask Group Title
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    the periodic table i am using says atomic weight =Al M=26g/mol is this the right thing to use as M

    • one year ago
  22. Jonask Group Title
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    174.85 s the answer thanks

    • one year ago
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