Question #bf7ca

Let's look at our reactants and products,

Reactants

• 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)#.

Products

• 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 #(aq)#.

• 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:

#"Zn"(s) + "AgNO"_3 (aq) rarr "Zn(NO"_3")"_2 (aq) + "Ag" (s)#

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 #2# before the silver nitrate on the left side:

#"Zn"(s) + color(blue)(2)"AgNO"_3 (aq) rarr "Zn(NO"_3")"_2 (aq) + "Ag" (s)#

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 #2# in front of the silver on the right side:

#"Zn"(s) + color(blue)(2)"AgNO"_3 (aq) rarr "Zn(NO"_3")"_2 (aq) + color(red)(2)"Ag" (s)#

Now, all the species are present equally on both sides of the reaction arrow, so this equation is balanced.