Let's look at our reactants and products,
- zinc metal. We write this in the chemical equation as the chemical symbol for zinc,
#"Zn"#. Often times a species in a chemical equation is followed by its physical state in parentheses:
#(s)#for a solid substance
#(l)#for a liquid substance
#(g)#for a gaseous substance
#(aq)#for a substance dissolved in aqueous solution
Most periodic tables will indicate the state of matter for each element, usually via a color code. Zinc is a solid metal, so we can follow it with the symbol
#(s)#in the reaction
- silver nitrate. This is a compound formed from the electrostatic attraction between a cationic silver atom,
#"Ag"^+#, and the anionic nitrate species, #"NO"_3^-#, so its formula is #"AgNO"_3#.
As for the state of matter, we must realize something about this reaction. Most all ionic substances are solid at room temperature, but for this reaction we won't write
#(s)#next to it, and here's the simple reason why. When two pure solids are interacting, nothing chemically is really happening; you can squeeze the two substances together, but no new substances will form. Therefore, it is implied that the #"AgNO"_3#species is dissolved in aqueous solution, so we follow it with #(aq)#.
- zinc nitrate. This compound is formed from the attraction between the zinc ion
#"Zn"^(2+)#and the nitrate ion, so its formula is #"Zn(NO"_3")"_2#.
Its state of matter is the same situation as the silver nitrate, and is dissolved in solution, so it's followed by
- silver metal. Likewise for zinc metal, this is simply the chemical symbol of silver,
#"Ag"#followed by its state of matter, solid, so it is #"Ag"(s)#
We can now form an unbalanced (no coefficients) chemical equation from this:
If you'll notice, there are two nitrate ions on the right side, and one on the left. To balance the nitrate species, we can place a
What you may notice now is that now the silvers are unbalanced; there are two on the left and one on the right. We can fix this by placing a
Now, all the species are present equally on both sides of the reaction arrow, so this equation is balanced.