1) Mass of first empty beaker (beaker 1) 2) Mass of beaker 1 + 200ml 0.2M AgNO3
ID: 861765 • Letter: 1
Question
1) Mass of first empty beaker (beaker 1)
2) Mass of beaker 1 + 200ml 0.2M AgNO3
3) Mass of beaker 1 + 200ml 0.2M AgNO3 + Cu
4) Mass of Cu used
5) Observations of the reaction
6) Mass of second empty beaker (beaker 2)
7) Mass of beaker 2 + Cu solution after reaction
8) Mass of beaker 1 + Ag solid after reaction
9) Mass of silver nitrate solution in beaker 1 = 3.395g of silver nitrate solution
10) mass of copper solution produced, beaker 2 = 202.838g of copper solution
11) Moles of copper used = .0157mol
12)Mass of silver produced by the reaction
13)Moles of silver produced by the reaction
14)Using moles of Copper and moles of Silver (there ratio) to write and balance the reaction between copper and silver (use your mole values, lowest whole numbers).
15)If the silver in the beaker contained water during your last weighing, how would this affect your results?
16) Assume that magnesium would act atom-for-atom exactly the same as copper in this experiment. How many grams of magnesium would, have been used in the reaction if one gram of silver were produced? The atomic mass of magnesium is 24.31 g/mol.
17) Explain the source of the blue color of the solution after the reaction.
I have 1-11 done..
Explanation / Answer
Hello,
Im kind of confused as to how your teacher wants you to find the answers, but I will try my best to help you out.
In this reaction Cu reduces Ag because Cu is higher on the acitivity series.
the balanced equation is,
2AgNO3 + Cu(s) --> Cu(NO3)2 + 2Ag(s)
Number 13:
You said that there are .0157 mols of Cu used so it is the limiting reactant because there was .04 mol of Ag in the solution, therefore only .0314 mols of Ag can be displaced. .0157 mol Cu x 2 mol Ag/ 1mol Cu = .0314
Number 12:
.0314 mol Ag x 107.87 g / mol = 3.39 g Ag produced
Number 14:
This is were I get confused becasue the equation is already balanced, so im not sure if you were susposed to come up with this ratio based on experimental values, but anyway the ratio is 2mol Ag / 1 mol Cu
Number 15:
Im assuming you extracted the solid silver from the experiment and if it was still wet it would weigh more, causing you to believe that there is a higher Ag to Cu ratio. This would result in an error when trying to write a properly balance the equation.
Number 16:
This states that Mg has a 2+ oxidation state, just as Cu. So,
1 g Ag x (1 mol Ag/107.87 g Ag) x (1mol Mg/ 2 mol Ag) x (24.31 g Mg / mol Mg) = .113 g
Number 17:
The blue color is due to the copper in solution
How this helps some, and sorry i was a little late answering it!
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