Warmup 12/2/16 Objective Tonight’s Homework What do you think it means when we say 2 chemicals can react? Objective Tonight’s Homework To learn how to count atoms before and after a reaction None

Notes on Reactants and Products 2SO2 + O2  2SO3    1) We call these reactants. They’re the things that are going to react. 2) The arrow tells us which direction the reaction will go. Most of the time it will face to the right. 3) These are the products. They’re the end result of what’s reacted. { { 2

Notes on Reactants and Products 2SO2 + O2  2SO3 When looking at chemical reactions, we have to make sure they’re balanced. This means we have to have the same number of each type of element on both sides.

Notes on Reactants and Products 2SO2 + O2  2SO3 When looking at chemical reactions, we have to make sure they’re balanced. This means we have to have the same number of each type of element on both sides. Let’s see if this reaction is balanced. On the left, we have 2 S atoms. On the right, we also have 2 S atoms. On the left, we have 4 O atoms + 2 O atoms. On the right, we have 6 O atoms.

Notes on Reactants and Products So what happens if the reaction isn’t balanced? We have to balance it. This can be tricky, but if you think about it the right way, it’s not. Let’s think of it like a recipe. Below is a recipe “equation”. 3 eggs + 1 flour + 1 sugar  1 cake

Notes on Reactants and Products So what happens if the reaction isn’t balanced? We have to balance it. This can be tricky, but if you think about it the right way, it’s not. Let’s think of it like a recipe. Below is a recipe “equation”. 3 eggs + 1 flour + 1 sugar  1 cake What if this wasn’t “balanced”? What if it really took 2 sugar to make? We could just change the “1” in front of sugar to a “2”.

Notes on Reactants and Products Chemical reactions work the same way. If we want to balance it, we can only change the number in front of each compound. We can’t change the small numbers that come after each element in a compound. That would change the compound into a different material. So to summarize. If you need to balance: 1) Count the number of each type of atom on the left side and compare it to the right side. 2) If they match, you’re done. If not, you need to change the numbers in front of each compound until they do.

Notes on Reactants and Products One last thing to note: Let’s say we have the following: 4 HCl + 4 Na → 4 NaCl + 2 H2 You’ll notice that we have numbers in front of each compound. Not only that, but we can divide everything by 2. What we have now would be like a recipe for 2 cakes. We only want 1. So if you can, divide all the numbers in front of each molecule by the same thing until you’re down to the lowest common denominator. 2 HCl + 2 Na → 2 NaCl + H2

Notes on Reactants and Products Example: ___ NaNO3 + ___ PbO  ___ Pb(NO3)2 + ___ Na2O

Notes on Reactants and Products Example: _2_ NaNO3 + ___ PbO  _1_ Pb(NO3)2 + ___ Na2O Let’s look at Pb(NO3)2. The little 2 tells us that we have 2 NO3 groups. If we want the left to balance, I’m going to need a 2 in front of NaNO3.

Notes on Reactants and Products Example: _2_ NaNO3 + ___ PbO  _1_ Pb(NO3)2 + _1_ Na2O So now we have 2 Na on the left. Looking at the right, if I put a 1 in front of Na2O, we’ll get 2 Na on the right, too.

Notes on Reactants and Products Example: _2_ NaNO3 + _1_ PbO  _1_ Pb(NO3)2 + _1_ Na2O Last is oxygen. On the right, I have: 6 for the Pb(NO3)2 And 1 from the Na2O. So I need 7 O on the left to balance. 2 NaNO3 gives me 6, so we need a 1 in front of the PbO to make it balance.

Notes on Reactants and Products Example: _2_ NaNO3 + _1_ PbO  _1_ Pb(NO3)2 + _1_ Na2O Lastly, double check everything. If we do, we find that everything on the left has the same totals as everything on the right. We can’t divide our numbers down, so it looks like we’re done! If you final check shows some things don’t balance, keep changing numbers until they do!

Exit Question ___ Fe + ___ AgNO3  ___ Fe(NO3)2 + ___ Ag How many silver nitrate are there in the final balanced equation? a) 0 b) 1 c) 2 d) 3 e) 4 f) None of the above