chapter9_epkd

= Chapter 9 - Stoichiometry = = //Note: scroll to bottom to find answers to examples with an astrix// = = Stoichiometry is just like a baking recipe. (ex:If a recipe calls for 2 eggs and 3 cups of flour, but you only have 1 egg, then you can only use 1.5 cups of flour to make the reicpe proportional). So, if the reacion (below) requires 8 moles of NaCl, then 8 moles of Na and 4 moles of Cl 2  are required to make the reaction possible. . = = = = = = = = = Equation Given: 2Na + Cl2 --> 2NaCl -multiply all elements by 4 (to get 8 moles of NaCl [you must multiply all, never just one]). Final Equation: 8Na + 4Cl2 --> 8NaCl

1) Write Equation 2) Balance Equation 3) Find Out How Many Moles are Necessary 4) Use Molar Mass to Convert to Grams** || //Equation Given:// Fe + O2 --> Fe2O3 //Balance:// 4Fe + 3O2 --> 2Fe2O3 //How Many Grams of Fe + O2 are Required to Make 10 Moles of Fe2O3?:// //Use Molar Mass to Convret to Grams:// 20 mol Fe x (55.845g/1mol Fe) = 1116.9 grams 15 mol O2 x (31.998g/1mol O2) = 479.9 grams
 * < **__Beginning Steps to Solving a Stoichiometry Problem:__
 * Finished Example:**

How many grams of H2 are required to make 75 grams of NH3 in the following reaction? N2 + H2 --> NH3
 * Complete Example By Using Beginning Steps:***

75g NH3 (1mol NH3/17.036g) x (3mol H2/2mol NH3) x (2.02g H2/1mol H20) = 13.3 grams of H2
 * Alternative Way to Complete Example (combine steps):**

//REMEMBER:// make sure all measurements balance out (i.e. grams on top and on bottom, etc.)

If you have 25.2 grams of N2 + 15.2 grams H2, is your limiting reactant in the production of NH3? 25.2 of N2 (1 mol/28g) ( 2 mol/1 mol) (17.03g/1 mol) = 30.654 g 15.2 (1mol/2.02g) (2 mol/3 mol) (17.03/1ol) = 85.431g //Limiting Reactant = 30.654g//
 * Limiting Reactant Problem:**

//To find the volume of a reactanct or product, just add this conversion to the end of the problem to make the final answer in liters.//
 * = **Mole to Volume Conversion** ||
 * = **1 mole** ||= **22.4 liters** ||

1) Find Individual Amount of Grams Per Element 2) Find Total Number of Grams 3) Create Proportions to Find Percentages** || //Given:// Ca3(PO4)2 //Individual Amount of Grams:// Ca3 = 120 (40 x 3) P = 62 (31 x 2) O4 = 128 (16 x 8) //Find Total Number of Grams:// 310 grams/mole //Create Proportions:// Ca3 = 38% (120/310) P= 26% (62/310) O4 = 42% (128/310) 1) Find the Number of Moles of Each Compound 2) Divide by the Smallest Number of Moles to Get a Ratio 3) Apply that Ratio to the Compound** ||
 * **__Steps to Find Percentage Composition:__
 * Finished Example:**
 * **__Steps to Find Emperical Formula:__
 * Empirical Formula - smallest whole number taio of atoms in a compound**

A compound is found to be 34.8 O and 65.2 g As by mass. Find the Emp. Form.
 * Finished Example:**

34.8g O (1 mol/ 16 g) = 2.17 mol 65.2g As (1 mol/75 g) = .87 mol compound: As2O5 ratio: 1:2.5 1) Find the Emp. Form. 2) Find the Mass of the Emp. Form. 3) Divide the Mass of the Molecular Mass by the Mass of the Emp. Form. 4) Multiply Your Emp. Form. by the Answer From Step 3** ||
 * **__Steps to Fine Molecular Formula:__
 * Molecular Formula - actual formula of a compound

Finished Example:** //Find Empirical Formula:// CH //Find Mass of Empirical Formula:// 13 grams/mole //Divide:// 78/13 = 6 //Multiply by Step 3:// C6H6

__**Answers to Example Problems:**__
 * 75g NH3 x (1mol NH3 / 17.036g) x (3mol H2 / 2mol NH3) x (2.02g H2 / 1mol H2) = 13.3 grams of H2

//__Works Cited: "Stoichiometry" 19 May 2009 Chem4kids__// [|//http://www.chem4kids.com/files/react_stoichio.html//]