Question #bf7ca
1 Answer
Explanation:
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 solutionMost 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:
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.
