In: Chemistry
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)
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