1. For the following, predict the products (if any). If there is no reaction, wr

Question

1. For the following, predict the products (if any). If there is no reaction, write “no rxn”. For the

questions where reaction occurs, write:

balanced molecular equation

balanced complete ionic equation

balanced net ionic equation

classify the reaction as either precipitation or acid base resulting in a gas

Include phases for all species.

a) NaI(aq) + Pb(ClO4)2(aq)

b) Ba(NO3)2(aq) + (NH4)2SO4(aq)

c) HCl(aq) + NaHSO3(aq)

d) Cu(CH3COO)2(aq) + Rb2CO3(aq)

e) aqueous potassium carbonate + hydrobromic acid

f) Mg(NO3)2(aq) + Zn(CH3COO)2(aq)

g) aqueous silver perchlorate + aqueous magnesium bromide

h) NH4Cl(aq) + KOH(aq)

Explanation / Answer

a). NaI(aq) + Pb(ClO4)2(aq)

Balanced reaction à    Pb(ClO4)2(aq) + 2 NaI(aq) = PbI2(s) + 2 NaClO4(aq)
Reaction type: double replacement

Balanced complete ionic reaction à Pb2+ + 2ClO4- + 2Na+ + 2I- PbI2 (s) + 2Na++2Clo4

Balanced net ionic reaction à To get to the net ionic equation, you need to start with the balanced complete ionic equation. For your reaction, you need to recognize that all of the substances except for PbI2 are soluble, and so in aqueous solution, they exist as independent ions.

So, you start with:
Pb2+(aq) + 2 ClO4-(aq) + 2 Na+(aq) + 2 I-(aq) -->
PbI2(s) + 2 Na+(aq) + 2 ClO4-(aq)

To get to the net ionic equation, you cancel anything that is in exactly the same form on both sides of the equation, which will leave you with:

Pb2+(aq) + 2 I-(aq) --> PbI2(s)

b). Ba(NO3)2(aq) + (NH4)2SO4(aq)

Balanced reaction à    Ba(NO3)2(aq) + (NH4)2SO4(aq) = BaSO4(s) + 2 NH4NO3(aq)
Reaction type: double replacement

Balanced complete ionic reaction à Ba2+ 2NO3 2NH4+ SO42-à Ba2+ SO42- + 2NH4+ NO32-

Balanced net ionic reaction à
Write the total equation

(NH4)2SO4 + Ba(NO3)2 = BaSO4 + NH4NO3

Balance

(NH4)2SO4 + Ba(NO3)2 = BaSO4 + 2NH4NO3

Now break up the compounds based on solubility rules, or use the original state of the material. For example in #14 Al(OH)3 appears as the solid in the equation.

2(NH4+) + SO4(-2) + Ba+2 + 2(NO3-) = BaSO4(s) + 2NH4+ 2NO3-
All but BaSO4 are (aq)

Eliminate the like species that appear on both sides

SO4(-2)(aq) + Ba+2(aq) = BaSO4(s)

that is your net ionic.

It will always depend on the formation of a solid, a gas, or a non-ionic species. You can also have the breakup of a solid or the reaction of a gas. Look for things produced or used that are not (a

c). HCl(aq) + NaHSO3(aq)

Balanced reaction à    NaHSO3(s) + HCl(aq) = NaCl(aq) + H2O(l) + SO2(g)

Balanced net ionic reaction à HSO3- + H+ = H2O + SO2(g)

Balanced complete ionic reaction àNa+ H+ + SO32- + H+à Na+ Cl- + 2H+ + [O]- + SO22-

d). Cu(CH3COO)2(aq) + Rb2CO3(aq)à no reaction

e). aqueous potassium carbonate + hydrobromic acid

Balanced reaction à    2HBr + K2CO3 --> 2KBr + H2O + CO2

Balanced net ionic reaction à K2CO3 (aq) + HBr (aq) KBr (aq) + CO2 (g) + H2O (l)

write the ions for the compounds that are soluble, or are solutions:

K^(+) (aq) + CO3^(2-) (aq) + H^(+) (aq) + Br^(-) (aq) K^(+) (aq) + Br^(-) (aq) + CO2 (g) + H2O (l)

cancel the spectator ions (the ones that are the same on both sides of the equation):

CO3^(2-) (aq) + H^(+) (aq) CO2 (g) + H2O (l)

balance:
CO3^(2-) (aq) + 2H^(+) (aq) CO2 (g) + H2O (l)

Balanced complete ionic reaction à2 H+ 2Br- + 2K+ CO32-à 2K+ Br- + 2H+ [O]- + CO2-

f). Mg(NO3)2(aq) + Zn(CH3COO)2(aq)à no reaction

g). aqueous silver perchlorate + aqueous magnesium bromide à no reaction

h). NH4Cl(aq) + KOH(aq)

Balanced reaction à NH4Cl(aq) + KOH(aq) = NH4OH(aq) + KCl(aq)
                                          Reaction type: double replacement

NH4Cl(aq) + KOH(aq) ---> KCl(aq) + NH4OH (aq)

No solid forms which means this does not precipitate and will not have a net equation because no reaction occurs

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.