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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
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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sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1Start by balancing the equation Cu+ 2AgNO3 > Cu(NO3)2 + 2Ag

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1Do you know how to determine the limiting reactant?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why are both of them in Ag?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so the limiting one would be the silver nitrate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's it?? That seemed kind of easy...I think...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So to see how many silver nitrate was formed I would add 8.4 and 6.3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm so confused right now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Srry no question mark I am confused

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1Alright imma restart this then try to make it clearer.

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1There 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 4.23.15 = 1.05 which is also 2.1*.5. So the amount of Cu left over is 1.05moles.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't get it nvm just for get it

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1Sorry I cant think of how to explain it very well...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its ok. Its just chem..
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