1. 4.06 Stoichiometry

2. It’s a process that allows us to mathematically convert and calculate the relationships between the amount of reactants and products in a chemical reaction.

3. Graphical representation Example: for every O2 molecule, there are 2 H2O molecules 2 H2 + 1 O2  2 H2O

4. Solving stoichiometry equations It is just like solving an usual math equation. Just to let you know, stoichiometry equations start with one substance and ends with another substance. 1. Balance the unequal equation by adding coefficients. 2.Identify what you have been asked to find in the question. 3.Set up your equation with the given information you have, which should be the measure and unit of a substance. 4.Solve by multiplying everything in the numerator and dividing by everything in the denominator. 5.Oh yeah, and don’t forget to check your work!

5. Show me Stoichiometry It is important that you always keep a periodic table handy while solving stoichiometry problems. You will also encounter stoichiometry questions that ask you for limiting reactants and percent yield, but don’t be intimidated. These are quite easy to calculate as well. Now lets solve a few.

6. Problem #1 How many moles of aluminum hydroxide are needed to react completely with 12.3 moles of nitric acid? Al(OH) 3 + 3 HNO 3 → 3 H 2 O + Al(NO 3 ) 3 Step 1: Balance the equation, which in this case the equation doesn’t need balancing because it already is. Step 2: Identify what the question asks and what you have been given. We are asked to find the amount of moles needed in order to react with 12.3 moles of nitric acid. 12.3 moles of nitric acid is our given information that we have to use to solve fore aluminum hydroxide. Step 3: Set up the equation with given information. We must use 12.3 moles of nitric acid to solve. Step 4: Solve. This is what the equation should look like. 12.3 mol HNO3 x 1 mol Al(OH)3 (We must divide by 3 mol HNO3 to get 3 mol HNO3 rid of nitric acid) 12.3 x 1/3 = 4.1 mol Al(OH)3 This is the final answer.

7. Problem #2 How many grams of carbon dioxide are produced when 46.8 grams of C2H2 combust? 2 C 2 H 2 + 5 O 2 → 4 CO 2 + 2 H 2 O Step 1: Balance the equation, which it does not need balancing. Step 2: Identify what you are asked to find and what you have been given(46.8 g of C2H2). Step 3: Set up an equation with 46.8 g of C2H2 as your starting amount. Step 4: Solve for carbon dioxide. The equation looks like this. 46.8 g C2H2 x 1 mol C2H2 x 4 mol CO2 x 44.01 g CO2 = 158. 2 g CO2 26.03g C2H2 2 mol C2H2 1 mol CO2 If you are wondering where the grams came from in the equation, they are the atomic mass of elements combined with other elements into one mass. We come up with the final answer bymultiplying everything in the numerator and dividing by everything in the denominator. 158.2 grams of carbon dioxide is the final answer when we have multiplied and divide everything.

8. Problem#3: examples, examples and more examples How many grams of chlorine gas can be produced when 62.3 grams of aluminum chloride decompose? 2 AlCl 3 →2 Al + 3 Cl 2 Step 1: Balance the equation. Step 2: Identify your given info and what you are asked to find(grams of chlorine gas and 62.3 grams of aluminum chloride). Step 3: Set up an equation with your given information. Step 4: Solve for grams of chlorine gas. 62.3 g AlCl3 x 1 mol AlCl3 x 3 mol Cl2 x 70.8g Cl2 = 49.7 g 133.3 g AlCl3 2 mol AlCl3 1 mol Cl2 Cl2 is the final answer