Precipitate Reactions
A precipate reaction is basically a chemical reaction that results with the formation of a solid of some sort.
It doesn't require too much math, luckily, but it does require a few add-ons, for example, a solubility chart.
The example we'll use is AgNO3(aq) + Na2S(aq). First off, we'd have to figure out what are the products using Ms. McRae's dance partner method - if Boy 1 (Ag) goes with Girl 2 (S), and Boy 2 (Na) goes with Girl 1 (NO3), we'd be left with our products being AgS, and NaNO3. Then, we have to balance the equation - sulfur has a charge of 2, so we need two silvers, and since there's already 2 sodiums, we need to 2 nitrates as well, leading to the final balanced chemical equation being: 2AgNO3 + Na2S --> Ag2S + 2NaNO3.
So, now there's two ways to figure out which is the precipitate - one using a solubility chart, where you would find the two substances that make up the compound, and find whether or not it's soluble. Here's a good example of a solubility chart: http://intro.chem.okstate.edu/1515sp01/database/solub.html
Your other option is to learn the solubility guidelines, which can be found right here:
http://www.fairbornchempage.com/Resources/SolGuides.htm
Either option is good for now, however you will be expected to have a general understanding of the guidelines this year, and to rely less on the solubility chart.
So, as you've probably already discovered, the answer to this question in the end is:
2AgNO3(aq) + Na2S(aq) --> Ag2S(s) + 2NaNO3(aq)
But, that's not the end of precipitate equations - there's still one last step: turning it into a net ionic equation.
This form exists, to show which reactants and products are just floating in the solution, and which of them actually react. The way you do this is fairly simple.
After going through these last three steps, what you're going to do, is take all the aqueous products - not the solids or gases, or even liquids - just aqueous.
Out of the aqueous products, you're going to find the electrolytes - any of the strong bases and acids, along with salts.
So, let's look at our example: 2Ag(+1)NO3(-1)(aq) + Na2(+1)S(-2)(aq) --> Ag2(+1)S(-2)(s) + 2Na(+1)NO3(-1)(aq)
AgNO3, Na2S, and NaNO3 are all strong electrolytes, so we immediately separate those, making the new equation:
2Ag(+1)(aq) + 2NO3(-1)(aq) + 2Na(+1)(aq) + S(-2)(aq) --> Ag2(+1)S(-2)(s) + 2Na(+1)(aq) + 2NO3(-1)(aq)
After that, you cancel out all the ions that appear more than once, which in this case would leave us with a final equation of...
2Ag(+1)(aq) + S(-2)(aq) --> Ag2(+1)S(-2)(aq)
And that, is how you write, a precipitate reaction!
It doesn't require too much math, luckily, but it does require a few add-ons, for example, a solubility chart.
The example we'll use is AgNO3(aq) + Na2S(aq). First off, we'd have to figure out what are the products using Ms. McRae's dance partner method - if Boy 1 (Ag) goes with Girl 2 (S), and Boy 2 (Na) goes with Girl 1 (NO3), we'd be left with our products being AgS, and NaNO3. Then, we have to balance the equation - sulfur has a charge of 2, so we need two silvers, and since there's already 2 sodiums, we need to 2 nitrates as well, leading to the final balanced chemical equation being: 2AgNO3 + Na2S --> Ag2S + 2NaNO3.
So, now there's two ways to figure out which is the precipitate - one using a solubility chart, where you would find the two substances that make up the compound, and find whether or not it's soluble. Here's a good example of a solubility chart: http://intro.chem.okstate.edu/1515sp01/database/solub.html
Your other option is to learn the solubility guidelines, which can be found right here:
http://www.fairbornchempage.com/Resources/SolGuides.htm
Either option is good for now, however you will be expected to have a general understanding of the guidelines this year, and to rely less on the solubility chart.
So, as you've probably already discovered, the answer to this question in the end is:
2AgNO3(aq) + Na2S(aq) --> Ag2S(s) + 2NaNO3(aq)
But, that's not the end of precipitate equations - there's still one last step: turning it into a net ionic equation.
This form exists, to show which reactants and products are just floating in the solution, and which of them actually react. The way you do this is fairly simple.
After going through these last three steps, what you're going to do, is take all the aqueous products - not the solids or gases, or even liquids - just aqueous.
Out of the aqueous products, you're going to find the electrolytes - any of the strong bases and acids, along with salts.
So, let's look at our example: 2Ag(+1)NO3(-1)(aq) + Na2(+1)S(-2)(aq) --> Ag2(+1)S(-2)(s) + 2Na(+1)NO3(-1)(aq)
AgNO3, Na2S, and NaNO3 are all strong electrolytes, so we immediately separate those, making the new equation:
2Ag(+1)(aq) + 2NO3(-1)(aq) + 2Na(+1)(aq) + S(-2)(aq) --> Ag2(+1)S(-2)(s) + 2Na(+1)(aq) + 2NO3(-1)(aq)
After that, you cancel out all the ions that appear more than once, which in this case would leave us with a final equation of...
2Ag(+1)(aq) + S(-2)(aq) --> Ag2(+1)S(-2)(aq)
And that, is how you write, a precipitate reaction!