Chemistry Question!? Net Ionic Equations

Write the net ionic equation only for any precipitation reaction which forms when the following solutions are mixed:

a) sodium sulphide and lead (II) nitrate

b) calcium hydroxide and aluminium iodide

c) barium hydroxide and magnesium sulphate

d) ammonium phosphate and calcium iodide

Please help, I am in an online course & the teacher is useless! I am so close to being finished and I greatly appreciate any help. Thanks!

Write the net ionic equation only for any precipitation reaction which forms when the following solutions are mixed:
The web site below has solubility tables to determine whether a compound will precipitate.
http://www.google.com/webhp?hl=en&tab=ww…

The 2nd image under “Images for chemistry solubility table” has compounds listed.
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All of these reactions are double replacement.
A+B- + C+D- → A+D- + C+B-
1) The products have the positive and negative ions switched
2) Balance the charges with subscripts, use parenthesis for polyatomic ions.
3) Balance the atoms with coefficients.
4) Check each product to determine solubility


a) sodium sulphide and lead (II) nitrate
1) NaS + PbNO3 → NaNO3 + PbS


2)
Na+1 S-2 → Na2S, Pb+2 NO3 -1 → Pb(NO3)2, Na+1 NO3 -1 → NaNO3, Pb+2 S-2 →PbS
2) Na2S + Pb(NO3)2 → NaNO3 + PbS

3) Need 2 Na on right side → 2NaNO3
Na2S + Pb(NO3)2 → 2NaNO3 + PbS
Atoms are balanced

4) PbS is insoluble, so it precipitates
Na2S + Pb(NO3)2 → 2NaNO3 + PbS (s)
(s) means solid (does not dissolve)

b) calcium hydroxide and aluminium iodide
CaOH + Al I → Ca I + Al OH
Ca +2 OH-1 = Ca(OH)2, Al+3 I-1 →Al I3, Ca+2 I -1 → Ca I2, Al+3 OH-1 → Al(OH)3

Ca(OH)2 + Al I3 → Ca I2 + Al(OH)3
Subscripts are 2’s and 3’s, so we need 6 OH and 6 I, use coefficient of 3 and 2
3 Ca(OH)2 + 2 Al I3 → 3 Ca I2 + 2 Al(OH)3
Al(OH)3 = s = insoluble
3 Ca(OH)2 + 2 Al I3 → 3 Ca I2 + 2 Al(OH)3 (s)


c) barium hydroxide and magnesium sulphate
Ba+2 OH-1 + Mg+2 SO4 -2 → Ba+2 SO4 -2 + Mg +2 OH-1
Ba(OH)2 + MgSO4 → BaSO4 + Mg(OH)2
2 OH = 2 OH
OK

d) ammonium phosphate and calcium iodide
NH4 +1 PO4 -3 + Ca+2 I -1 → NH4 +1 I-1 + Ca+2 PO4 -3
(NH4)3PO4 + Ca I2 → NH4 I + Ca3(PO4)2
Need 3 Ca on left side = 3 Ca I2

(NH4)3PO4 + 3 Ca I2 → NH4 I + Ca3(PO4)2
Need 6 I on right side = 6 NH4 I

(NH4)3PO4 + 3 Ca I2 → 6 NH4 I + Ca3(PO4)2
need 6 NH4 and 2 PO4 on left side = 2 (NH4)3PO4

2 (NH4)3PO4 + 3 Ca I2 → 6 NH4 I + Ca3(PO4)2
Ca3(PO4)2 = s = precipitate
2 (NH4)3PO4 + 3 Ca I2 → 6 NH4 I + Ca3(PO4)2 (s)

I hope this helps