NAME THAT PRODUCT - PREDICTING THE PRODUCTS OF CHEMICAL REACTIONS

ANSWER KEY

The products are shown for each chemical reaction.  An X is placed in the correct cell in the tables below each reaction. 

1.  Na₂SO₄(aq) + Pb(NO₃)₂(aq) à 2NaNO₃(aq) + PbSO₄(s)
PRECIPITATION RX - mixing 2 aqueous salt solutions. With Pb, think precipitation.  The spectator ions are sodium and nitrate.

2.  2C₂H₆(g) + 7O₂(g) à 4CO₂(g) + 6H₂O(g)

REDOX RX - pure element in reactants (O₂) disappears into compound of oxygen; ox. # of O on left is 0, on right -2, implying that O is reduced and acts as an oxidizing agent. The average Ox. # of C in C₂H₆ is 3- (each H is 1+ for total of 6+ so avg C is 3-). Ox. # of C in CO₂ is 4+. Hence C is oxidized and is a reducing agent.  Clearly redox reaction. No simple proton donor or acceptor, and no aqueous solutions being mixed.

3.  2NaOH(aq) + MnSO₄(aq) à Na₂SO₄(aq) + Mn(OH)₂(s)

PRECIPITATION RX - mixing two aqueous salt solution in which each dissociated into ions. Even though NaOH is a strong base, there is not proton donor with which to react. The strong base is the source of OH^- ions which precipitate with Mn ions. Clearly not redox. Remember most sulfates are soluble except salts of calcium barium, lead, mercury, and silver. Spectator ions are sodium and sulfate.

4.  2HCl(aq) + Pb(NO₃)₂(aq) à 2HNO₃(aq) + PbCl₂(s)

PRECIPITATION RX - mixing two aqueous solutions in which each dissociated into ions. One might argue that this is an acid/base reaction as well since the strong acid HCl is donating a proton to a weak base NO₃ to form the strong acid HNO₃. However, nitrate is a spectator ion and doesn't participate in the reaction. The "driving force" for the reaction is the formation of the precipitate The strong acid is the source of Cl^- ions. Clearly not a redox. Cl on both sides has an ox. # of 1-.  Spectator ions are hydrogen (hydronium) and nitrate.

5.  3 HNO₃(aq) + Al(OH)₃(s) à  Al(NO₃)₃(aq) + 3H₂O(l)

ACID/BASE RX - HNO₃ is nitric acid, a strong acid and a strong electrolyte. Is there anything around to protonate or possibly precipitate? Clearly there is - Al(OH)₃(s). It is a solid but presumably would dissolve and the OH^- -  a strong base - reacts with the nitric acid and is protonated by the strong acid HNO₃. It can't be a precipitation reaction since start with a solid reactant. Isn't a redox rx like in lab 3 - HNO₃ + Cu(s), in which pure Cu dissolved and reacted to form Cu²⁺ ions. Al already exists as Al³⁺ ions.

6.  Ca(OH)₂(s) + 2HCl(aq) à 2H₂O(l) + CaCl₂(aq)

ACID/BASE RX - Have a strong acid (HCl) and a strong base (Ca(OH)₂) which is sparingly soluble. All Cl^1- salts except Ag, Hg, and Pb are soluble. No redox reaction when have obvious strong acid and base. 

7.  Cu(s) + 2AgNO₃(aq) à Cu(NO₃)₂(aq) + 2Ag(s)

REDOX RX - Clearly no acid or base; don't have two aqueous salt solutions to mix. Ox. # of Cu goes from 0 to 2+ (hence copper is oxidized and is a reducing agent) while Ag goes from 1+ to 0 (hence silver ion is reduced and is an oxidizing agent). Pure element disappear into compounds or form from a compound of the element.  This reaction would go only if Cu and Ag⁺ are better reducing agent and oxidizing agent, respectively, than the appropriate products, as we will  in the activity series lab.

8.  Zn(s) + 2HCl(aq) à ZnCl₂(aq) + H₂(g)

REDOX RX - This is from Lab 3. Excess Zn(s) dissolves in HCl and is oxidized to Zn²⁺ ions - hence it is a reducing agent). Even though one of the reactants is a strong acid, there is clearly no proton acceptor to accept. Likewise, don't have two aqueous ionic solutions mixing to form a solid. Zn ox. # goes form 0 to 2+ (hence it is oxidized and acts as a reducing agent) while H in HCl goes from 1+ to 0 (so it gets reduced and acts as oxidizing agent).  Zn is higher in the activity series than H₂ and hence displaces it from a compound of H. 

9.  H₂SO₄(aq) + BaCl₂(aq) à BaSO₄(s) + 2HCl(aq)

PRECIPITATION RX - Even though one of the reactants is a strong acid (H₂SO₄), it also is a source of ions that can form a precipitate with Ba²⁺. In this reactions, two ionic solutions are being mixed to form a precipitate. Solubility rules show that BaSO₄ is insoluble in water. One could argue that this is an acid base reaction as well since the acid (proton donor) is donating protons to a very weak base Cl^- to form HCl. However, Cl ion is a spectator ion and doesn't participate in the reaction. The "driving force" for the reaction is the formation of the precipitate.

10.  H₂SO₄(aq) + 2KOH(aq) à K₂SO₄(aq) + 2H₂O(l)

ACID/BASE RX - In this rx, a strong acid reacts with a strong base. Both acid and base are strong electrolytes in water. There is no change in oxid. # from reactants to products. Likewise, there is no precipitate formed.

11.  C₆H₁₂O₆(s-glucose) + 6O₂(g) à 6CO₂(g) + 6H₂O(g)

REDOX RX - This is an example of the oxidation of a carbohydrate with oxygen to form carbon dioxide and water - a reaction that goes on within us as you read this. A pure element (O₂) disappears to form a compound of that element. The oxid # of O in O₂ is O and goes to 2- in CO₂ and H₂O - hence it is reduced and acts as an oxidizing agent. The average oxidation number of C in C₆H₁₂O₆ is 0, since each H is 1+ and each O is 2-. The oxid # of C in CO₂ is 4+. Hence it is oxidized.   Clearly there is not precipitation reaction and no simple proton transfer.à

12. FeCl₃(aq) + 3KOH(aq) à Fe(OH)₃(s) + 3KCl(aq)

PRECIPITATION RX - Even though KOH is a strong base, there is no proton donor to interact with the OH^-. Therefore it is not an acid/bse reaction. Solubility rules show that Fe(OH)₃ should be insoluble. When you hear hoofsteps, don't think zebras!  Potassium and choride are spectator ions.

13.  NaNO₃(aq) + HCl(aq) à HNO3(aq) + NaCl(aq)

NO REACTION - All sodium salts are soluble; all nitrates are soluble. One could argue that it is an acid/base reaction, where HCl donates a proton to the weak base NO₃^- to form HNO₃. The problem with that is that HNO₃ is a strong acid itself, which completely dissociates. Therefore, NO₃^- is a spectator ion throughout. Nothing changes. Everything is a strong electrolyte.

14.   2Na₃PO₄(aq) + 3Pb(NO₃)₂ à Pb₃(PO₄)₂(s) + 6NaNO₃(aq)

PRECIPITATION RX - mixing 2 aqueous salt solutions. No acids/base around.  When you see Pb, think precipitate.   Sodium and nitrate are spectator ions.

15.  2Mg(s) + O₂(g) à 2MgO(s)

REDOX RX. - Clearly this can't be an acid/base or precipitation reaction. Two pure elements disappear into a compound of that element. Pure elements have an oxidation number of 0.  Mg is oxidized by the oxidizing agent oxygen.

16.   2 NH₄Cl(aq) + K₂SO₄(aq) à (NH₄)₂ SO₄(aq) + 2KCl(aq)

NO REACTION - Two aqueous salt solutions mixed together, but both new products are soluble. All ions are spectator ions/

17.  HF(aq) + KOH(aq) à H₂O(l) + KF(aq)

ACID/BASE RX - KOH is a strong base, and HF (similar to HCl) must be a proton donor or acid. No change in oxidation number occurs, and clearly no precipitation reaction occurs

The following reactions are from the copper experiment:  They are already balanced for you. Click on the links to see videos of the reactions you did in lab!

18.  Cu(s) + 4HNO₃(aq) --> Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

REDOX RX - Clearly, a pure element, Cu(s) disappears into a compound of that element.  The oxidation number of N changes from 5+ in HNO₃ to 4+ in NO₂.  Notice the 2 of the 4 nitrates from the reactant side remain as products in copper nitrate.  Its hard to categorize this reaction with respect to a composition change.  In part, Cu displaces NO₂ from its compounds, so its a displacement.  However part of the nitrate from nitric acid remains, so its more complicated than that. 

19.  Cu(NO₃)₂(aq) + 2NaOH(aq) ---> Cu(OH)₂(s) + 2NaNO₃(aq)

PRECIPITATION RX - This is a classic case of two aqueous salt solutions which form an insoluble salt.  Remember, most hydroxides are insoluble.  It can't be a acid/base reaction since there is not proton donor.

20.  Cu(OH)₂(s) ^heat > CuO(s) + H₂O(l)

ACID/BASE RX - This is a bit tricky, since where is the acid in the reactants?  The base is there - OH.  Notice, however, that there are two OH's in the reactant.  Remember that an acid and base react to form another acid and base.   Look at the products and consider the reverse reaction.  An O^2- from copper oxide, which is clearly a base, reacts with water as an acid, which donates a proton to the oxide, to form 2 OH 's - one from oxide gaining a proton and one from water which loses a proton.  Now consider the forward reaction again.  One OH acts as an acid and becomes oxide, while one acts as a base and becomes water.  It's not a precipitation reaction since you start with a solid and end with a solid.

21.  CuO(s) + H₂SO₄ ---> CuSO₄(aq) + H₂O(l)

ACID/BASE RX - Clearly, a strong acid (sulfuric acid) reacts with a strong base (copper oxide) to form a salt and water.  Notice that the base does not have to be soluble in water for it to react with the strong acid.

22.  CuSO₄(aq) + Zn(s) ---> Cu(s) + ZnSO₄(aq)

REDOX - Here, a pure element (Zn, oxidation number of 0) disappears into a compound of that element (ZnSO4 - oxidation number of Zn = 2+).  Zn is oxidized and acts as an reducing agent.  Likewise, the pure element Cu appears from a compound of copper as Cu²⁺ is reduced. 

23.  Zn(s)  +  HCl(aq) ---> ZnCl₂(aq) + H₂(g)

same as Rx 8 above.