anonymous
  • anonymous
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. @taramgrant0543664 @photon336
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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taramgrant0543664
  • taramgrant0543664
So the first step is to balance the equation are you able to do that?
anonymous
  • anonymous
I think.... Cu+2AgNO3 --> Cu(NO3)2 +2Ag
taramgrant0543664
  • taramgrant0543664
Perfect! So now we need to determine which one either the Cu or the AgNO3 is the limiting reagent. Do you happen to know which one it is?

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More answers

anonymous
  • anonymous
No would it be Cu?
taramgrant0543664
  • taramgrant0543664
So to determine which one is the limiting reagent we know that the ratio of Cu to Ag is 1:2 and the ratio of AgNO3 to Ag is 1:1 so 4.2 moles of Cu makes 8.4 moles of Ag and we have 6.3 moles of AgNO3 which makes 6.3 moles of Ag so our limiting reagent is actually the AgNO3
anonymous
  • anonymous
Wait how do you know the ratio?
taramgrant0543664
  • taramgrant0543664
the ratio is found with the stoichometric coefficients so when you balanced it you are finding the ratio
anonymous
  • anonymous
So whats next?
anonymous
  • anonymous
So would NO be worth nothing?? Since the ratio is 1:1 and 6.3 is Ag by itself too? @taramgrant0543664
taramgrant0543664
  • taramgrant0543664
So the next step since we have our limiting reagent as AgNO3 we will see how much of the Ag is produced Since we have found that AgNO3 and Ag are in the 1:1 ratio 6.3moles of Ag are produced from this reaction
taramgrant0543664
  • taramgrant0543664
The ratio includes NO as part of the AgNO3 compound this reaction is a single displacement reaction so the NO3 transfers to the Cu leaving the Ag by itself
taramgrant0543664
  • taramgrant0543664
Do you have everything so far?
anonymous
  • anonymous
so i would subtract 6.3 from 8.4?
taramgrant0543664
  • taramgrant0543664
Not exactly so we know that 6.3 moles of Ag are produced in this reaction and the ratio of Cu to Ag is 1:2 so we have to divide the 6.3 by two getting 3.15 and then we can subtract the 4.2-3.15= the amount left over
taramgrant0543664
  • taramgrant0543664
The reason why we don't use the 8.4 moles because that is us saying how much could be made from the Cu but we used the AgNO3 as our limiting agent
anonymous
  • anonymous
@taramgrant0543664 So it would be 1.05grams/mol of Cu?
taramgrant0543664
  • taramgrant0543664
It would be 1.05 moles of Cu not the 1.05g/mol
anonymous
  • anonymous
oh ok thx
taramgrant0543664
  • taramgrant0543664
No problem!

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