Answer:
[tex]5\; \rm mol[/tex] of [tex]\rm Cu(NO_3)_2[/tex] would be produced (assuming that reaction does not run out of [tex]\rm AgNO_3[/tex] until all the [tex]\rm Cu[/tex] was converted.)
Explanation:
Make sure that the equation for this reaction is indeed balanced:
[tex]\rm Cu + 2\, AgNO_3 \to Cu(NO_3)_2 + 2\, Ag[/tex].
The coefficient of [tex]\rm Cu[/tex] and [tex]\rm Cu(NO_3)_2[/tex] are not shown. That implies that the coefficient of both species would be [tex]1[/tex]. In other words, the actual equation for this reaction should be:
[tex]\rm 1\; Cu + 2\; AgNO_3 \to 1\; Cu(NO_3)_2 + 2\; Ag[/tex].
Coefficient of [tex]\rm Cu[/tex] in this equation: [tex]1[/tex].
Coefficient of [tex]\rm Cu(NO_3)_2[/tex] in this equation: [tex]1[/tex].
Thus, the ratio between the coefficient of [tex]\rm Cu(NO_3)_2[/tex] and that of [tex]\rm Cu[/tex] would be [tex](1 / 1 ) = 1[/tex].
Assume that [tex]\rm Cu[/tex] is the limiting reactant of this reaction (that is: this reaction runs out of [tex]\rm Cu\![/tex] before running out of any other reactant.) This coefficient ratio would be equal to the ratio between:
Hence, under the assumption that [tex]\rm Cu[/tex] is the limiting reactant:
[tex]\displaystyle \frac{n(\rm Cu(NO_3)_2)}{n(\rm Cu)} = \frac{\text{coefficient of $\rm Cu(NO_3)_2$}}{\text{coefficient of $\rm Cu$}} = \frac{1}{1} = 1[/tex].
The question states that [tex]5\; \rm mol[/tex] of [tex]\rm Cu[/tex] was available. That is: [tex]n({\rm Cu}) = 5\; \rm mol[/tex]. Assume that [tex]\rm Cu\![/tex] is indeed the limiting reactant.
[tex]\begin{aligned} n({\rm Cu(NO_3)_2}) &= \frac{n(\rm Cu(NO_3)_2)}{n(\rm Cu)} \cdot n(\rm Cu) \\ &= 1 \times (5\; \rm mol) = 5\; \rm mol\end{aligned}[/tex].
Hence, under this assumption, [tex]5\; \rm mol[/tex] of [tex]\rm Cu(NO_3)_2[/tex] would be produced.
