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anonymous

  • one year ago

If 4.2 moles of copper metal reacts with 6.3 moles of silver nitrate, how many moles of silver metal can be formed, and how many moles of the excess reactant will be left over when the reaction is complete? Unbalanced equation: Cu + AgNO3 → Cu(NO3)2 + Ag Be sure to show all of your work. @photon336 @welshfella @mathmate

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  1. sweetburger
    • one year ago
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    Start by balancing the equation Cu+ 2AgNO3 --> Cu(NO3)2 + 2Ag

  2. anonymous
    • one year ago
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    ok

  3. sweetburger
    • one year ago
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    Do you know how to determine the limiting reactant?

  4. anonymous
    • one year ago
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    no

  5. anonymous
    • one year ago
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    Why are both of them in Ag?

  6. anonymous
    • one year ago
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    Ok so the limiting one would be the silver nitrate

  7. anonymous
    • one year ago
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    That's it?? That seemed kind of easy...I think...

  8. anonymous
    • one year ago
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    So to see how many silver nitrate was formed I would add 8.4 and 6.3?

  9. anonymous
    • one year ago
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    I'm so confused right now?

  10. anonymous
    • one year ago
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    Srry no question mark I am confused

  11. sweetburger
    • one year ago
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    Alright imma restart this then try to make it clearer.

  12. anonymous
    • one year ago
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    ok

  13. sweetburger
    • one year ago
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    There is a 1:2 ratio between our Cu and AgNO3. If we have 4.2moles AgNO3 its a 2:1 ratio so 8.4moles. We only have 6.3moles of Cu though so 6:3:8.4 isnt 1:2 so 6.3moles is our limiting reagent and 6.3moles of product are formed so we need to find what is 1/2 of 6.3moles and that is 3.15moles so we had 3.15moles of Cu react. So we initiall had 4.2moles of Cu so 4.2-3.15 = 1.05 which is also 2.1*.5. So the amount of Cu left over is 1.05moles.

  14. anonymous
    • one year ago
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    I don't get it nvm just for get it

  15. sweetburger
    • one year ago
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    Sorry I cant think of how to explain it very well...

  16. anonymous
    • one year ago
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    Its ok. Its just chem..

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