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shaqadry Group Title

Consider the reaction between iron (III) oxide, Fe2O3 and carbon monoxide, CO. Fe2O3 + 3CO --> 2Fe + 3CO2 In the process, 213 g of Fe2O3 are reacted with 140 of CO. (a) calculate mass of Fe formed (b) how many moles of the excess reagent is left at the end of the reaction? Pls show me how to calculate this using dimensional analysis :(

  • 2 years ago
  • 2 years ago

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  1. Kryten Group Title
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    ok so you already have balanced chemical equation... so next step would be to write down what you have and what you need: Fe2O3 + 3CO --> 2Fe + 3CO2 m(Fe2O3)=213 g m(CO)=140 g --------------- n(Fe2O3)=? m(Fe)=? n(Fe2O3)=? n(CO)=? n(CO2)=? so now all you need to do is to get to it... n(Fe2O3)=m(Fe2O3) / M(Fe2O3) n(Fe2O3)= 213 g / 159,7 gmol-1 = 1,33 mol n(CO)= m(CO) / M(CO) n(CO)= 140 g / 28,01 gmol-1 = 4,99 mol now lets see what's the ratio in which reactants react to give products... (I have to go out for cca 15 min and when i get back i will finish this...)

    • 2 years ago
  2. shaqadry Group Title
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    i dont understand.. :(

    • 2 years ago
  3. Kryten Group Title
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    im back, can you tell me which part you dont understand so i can clear it up...

    • 2 years ago
  4. shaqadry Group Title
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    how do i calculate the limiting reactant? im very much confuse with everything now.

    • 2 years ago
  5. Kryten Group Title
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    we are coming to that in next step...

    • 2 years ago
  6. shaqadry Group Title
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    okay, go on.

    • 2 years ago
  7. Kryten Group Title
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    ok so now you need to get limiting reagent... to me it is obvious but that comes from experience, now how do you confirm which one is limiting reactant? you can calculate for any product what amount would you get from all reactants with chemical reaction ratios in mind... this sounds a bit confusing but let me explain by calculation and i hope you will get it... from reaction you get ratios (example 1): n(Fe) / n(Fe2O3) = 2/1 = 2 n(Fe) = 2 * n(Fe2O3) n(Fe) = 2 * 1,33 mol = 2,66 mol and n(Fe) / n(CO) = 2/3 n(Fe) = 2/3 * n(CO) n(Fe) = 2/3 * 4,99 mol = 3,32 mol now you can see that limiting reagent is Fe2O3 cause you get less product! now lets do second example (example 2): n(CO2) / n(Fe2O3) = 3/1 = 3 n(CO2) = 3 * n(Fe2O3) n(CO2) = 3 * 1,33 mol = 3,99 mol and n(CO2) / n(CO) = 3 / 3 = 1 n(CO2) = n(CO) n(CO2) = 4,99 mol and here too you get that Fe2O3 is limiting reagent cause less product is produced! do you get it?

    • 2 years ago
  8. shaqadry Group Title
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    i get it! so to calculate the limiting reactant i just go directly from the mass of the reactant to the mol of the product? is that correct?

    • 2 years ago
  9. Kryten Group Title
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    yes and to calculate limiting reagent you ALWAYS and i mean ALWAYS calculate with MOLES!!! not with mass or concentration but with moles! :D

    • 2 years ago
  10. Kryten Group Title
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    do you know how to continue?

    • 2 years ago
  11. shaqadry Group Title
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    okay thnks! and how do i calculate the excess?

    • 2 years ago
  12. Kryten Group Title
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    now that you know that Fe2O3 is limiting reagent you know that excess will be CO and you just calculate with Fe2O3 how much product will there be and reverse the process, calculate from that amount that you got from Fe2O3 how much CO is needed and substract from 4,99 moles... get it?

    • 2 years ago
  13. shaqadry Group Title
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    no...

    • 2 years ago
  14. shaqadry Group Title
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    you mean calculate the mole of CO using mole of Fe2O3?

    • 2 years ago
  15. shaqadry Group Title
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    then minus it with the originally mole of CO obtain?

    • 2 years ago
  16. Kryten Group Title
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    yes and no, calculate moles of CO2 and Fe from Fe2O3 since Fe2O3 is limiting reagent and from that moles of for example CO2 you calculate how much CO is needed. you already have moles of CO2 since we calculated it for limiting reagent! n(CO2) = 3,99 mol now you calculate reversing sides like this: n(CO) / n(CO2) = 3/3 = 1 n(CO) = n(CO2) n(CO) = 3,99 mol and excess is: n(CO, excess) = 4,99 mol - 3,99 mol = 1 mol get it?

    • 2 years ago
  17. shaqadry Group Title
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    yes i think i get it...

    • 2 years ago
  18. shaqadry Group Title
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    thank you!

    • 2 years ago
  19. Kryten Group Title
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    not a problem, glad to help... ;)

    • 2 years ago
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