Nitai M. answered 12/08/19
Medical student with 7+ years experience teaching chemistry
This is a double replacement reaction:
Al2(SO4)3(aq) + AgNO3(aq) → Al(NO3)3 + Ag2SO4
Now we must ask which of these products is soluble (and will dissociate into ions), and which are insoluble (and will remain solid). We will denote this by writing (aq) after any soluble species and (s) after each insoluble species.
Al(NO3)3 (aq) -- most nitrates are soluble, and this guy is no exception.
Ag2SO4 (s) -- most Ag salts are insoluble, and although many sulfates are soluble, this guy is insoluble.
Therefore, we can now include these phases in our equation and balance:
1Al2(SO4)3(aq) + 6AgNO3(aq) → 2Al(NO3)3(aq) + 3Ag2SO4(s)
Next, we dissociate the aqueous species into their ions, leaving solid compounds intact:
2Al3+(aq) + 3SO42-(aq) + 6Ag+(aq) + 6NO3-(aq) → 2Al3+(aq) + 6NO3-(aq) + 3Ag2SO4(s)
**Don't forget to multiply the balancing coefficient by the subscript coefficients whenever necessary. (For example, 2Al(NO3)3 -- we must remember to multiply the 3 NO3- ions per compound by the balancing coefficient, 2, to obtain the total number of NO3- ions participating in the reaction.)
Finally, we simplify by canceling species that appear on both sides of the equation:
2Al3+(aq) + 3SO42-(aq) + 6Ag+(aq) + 6NO3-(aq) → 2Al3+(aq) + 6NO-3(aq) + 3Ag2SO4(s)
We are left with the following net reaction:
3SO42-(aq) + 6Ag+(aq) → 3Ag2SO4(s)