This problem is asking for the consumption of silver ions when silver nitrate is reacted with copper. In such a case, since no masses are given, we can use the following from similar problems:
Mass of empty beaker: 110.000 g
Mass of beaker with silver nitrate (after all additions) and copper: 331.634 g.
Mass of beaker with silver: 113.395 g.
This means we can write the following chemical equation:
[tex]Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag[/tex]
And thus calculate the mass of silver nitrate that will produce the following mass of silver:
[tex]m_{Ag}^{produced}=113.395g-110.000g=3.395g[/tex]
Next, we use the 2:2 mole ratio of silver to silver nitrate (silver ions source):
[tex]3.395gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg} *\frac{169.87gAgNO_3}{1molAgNO_3} = 5.35gAgNO_3[/tex]
The step will be defined for the given mass of available silver nitrate which will be compared to 5.35 g (consumed mass) to see if they are the same (all consumed) or different (partial consumption).
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