Suppose you have \(3.00 \mathrm{~g}\) of powdered zinc metal, \(3.00 \mathrm{~g}\) of powdered silver metal and \(500.0 \mathrm{~mL}\) of a \(0.2 \mathrm{M}\) copper(II) nitrate solution. (a) Which metal will react with the copper(II) nitrate solution? (b) What is the net ionic equation that describes this reaction? (c) Which is the limiting reagent in the reaction? (d) What is the molarity of \(\mathrm{Cu}^{2+}\) ions in the resulting solution?

To determine which metal will react with Cu(NO₃)₂, we need to consult the activity series of metals. The activity series is a list that ranks metals based on their tendency to undergo redox reactions. A metal higher in the series will displace a metal ion lower in the series from its salt in an aqueous solution. The activity series for some common metals is: Mg > Al > Zn > Cr > Fe > Co > Ni > Sn > Pb > H > Cu > Ag Zinc (Zn) is higher in the activity series than copper (Cu), while silver (Ag) is lower. Therefore, zinc will react with the copper(II) nitrate solution, while silver will not.

The reaction between zinc and copper(II) nitrate is a single displacement reaction, in which, zinc displaces copper from its nitrate salt as follows: \(Zn(s) + Cu(NO_3)_2(aq) \rightarrow Zn(NO_3)_2(aq) + Cu(s)\) Now, let's write the net ionic equation for the reaction. First, separate all the soluble compounds into ions: \([Zn(s) \rightarrow Zn_{(s)}]\) \([Cu(NO_3)_2(aq) \rightarrow Cu^{2+}_{(aq)}+2NO_3^{-}_{(aq)}]\) \([Zn(NO_3)_2(aq) \rightarrow Zn^{2+}_{(aq)}+2NO_3^{-}_{(aq)}]\) Since nitrate ions are present on both sides, they are spectator ions and can be eliminated from the equation: Net ionic equation: \(Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}\)

We are given 3.00 g of zinc and a 500.0 mL solution of 0.2 M copper(II) nitrate. To determine the limiting reagent, we first need to calculate the moles of each reactant. Moles of zinc: \(\frac{3.00 \ \mathrm{g}}{65.38 \ \mathrm{g\ mol^{-1}}}=0.0459 \ \mathrm{mol}\) (using the molar mass of zinc) Moles of copper(II) nitrate: \(\mathrm{0.2 \ M} \times \mathrm{0.500 \ L} = 0.100 \ \mathrm{mol}\) Now, compare the stoichiometry of the balanced reaction: \(1 \ \mathrm{mol} \ Zn : 1 \ \mathrm{mol} \ Cu(NO_3)_2\) \(0.0459 \ \mathrm{mol} \ Zn : 0.100 \ \mathrm{mol} \ Cu(NO_3)_2\) Since there is less zinc than copper(II) nitrate in terms of their stoichiometry, zinc is the limiting reagent.

To calculate the molarity of the remaining Cu²⁺ ions in the solution, we first need to find the moles of Cu²⁺ ions that did not participate in the reaction. Moles of reacted Cu²⁺ ions: \(0.0459 \ \mathrm{mol}\) (based on the moles of limiting reagent – zinc) Moles of unreacted Cu²⁺ ions: \(0.100 \ \mathrm{mol} - 0.0459 \ \mathrm{mol} = 0.0541 \ \mathrm{mol}\) Now, we can calculate the molarity of Cu²⁺ ions in the resulting solution: Molarity of Cu²⁺ ions: \(\frac{0.0541 \mathrm{\ mol}}{0.500 \mathrm{\ L}} = 0.108 \mathrm{M}\) In conclusion: a) Zinc will react with the copper(II) nitrate solution. b) The net ionic equation for the reaction is: \(Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}\) c) The limiting reagent is zinc. d) The molarity of Cu²⁺ ions in the resulting solution is 0.108 M.