Question #c20f5

The easiest way to balance this chemical equation is to go by ions, not by atoms.

If you write the complete ionic equation for this reaction, you'll get

#3NH_(4(aq))^(+) + PO_(4(aq))^(3-) + Ca_((aq))^(2+) + 2Cl_((aq))^(-) -> Ca_3(PO_4)_(2(s)) + NH_text(4(aq])^(+) + Cl_((aq))^(-)#

If you were to eliminate spectator ions, i.e. the ions that are present on both sides of the equation, you'll get the net ionic equation, which looks like this

#PO_(4(aq))^(3-) + Ca_((aq))^(2+) -> Ca_3(PO_4)_(2(s))#

Notice that you need 3 calcium atoms on the products' side, but only have 1 calcium cation of the reactants' side #-># multiply the calcium cation by 3.

Likewise, multiply the phosphate anions by 2 to get them to match the number of phosphate ions present on the products' side.

This will get you

#color(red)(2)PO_(4(aq))^(3-) + color(blue)(3)Ca_((aq))^(2+) -> Ca_3(PO_4)_(2(s))#

Now take these stoichiometric coefficients and use them in the overall equation

#color(red)(2)(NH_4)_3PO_text(4(aq]) + color(blue)(3)CaCl_(2(aq)) -> Ca_3(PO_4)_(2(s)) + NH_4Cl_((aq))#

Now you only have to balance the ammonium ions, #NH_4^(+)#, and the #Cl#. Notice that you have 6 of each on the reactants' side, so multiply the ammonium chloride by 6 to balance everything out

#color(red)(2)(NH_4)_3PO_text(4(aq]) + color(blue)(3)CaCl_(2(aq)) -> Ca_3(PO_4)_(2(s)) + color(green)(6)NH_4Cl_((aq))#