STOICHIOMETRY

I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

What is Stoichiometry? Stoichiometry is the quantitative study of reactants an d products in a chemical reaction.

What You Should Expect To Do? Given : Amount of reactant(s) Question: How much product(s) can be formed? Example X A + Y B ? C “Given X grams of A and Y grams of B, how many grams of C can be produced?”

What Skills Do You Need? Some skills you will need to use: Molar Ratios (come from coefficients of a BALANCED equation). Molar Masses (Formula Weight calculations). Balancing Equations Conversions between grams and moles Note: This type of problem is often called "mass-mass."

Steps Involved in Solving Mass- Mass Stoichiometry Problems Balance the chemical equation correctly. Convert GRAMS to MOLES of the starting substance. Set up a MOLE RATIO between starting and ending substances. Convert MOLES to GRAMS of ending substance.

Mole Ratios A mole ratio COMPARES moles of one compound in a balanced chemical equation to moles of another compound (look at their coefficients).

Example In the reaction between magnesium and oxygen to form magnesium oxide, what is the mole ratio for the substances? 2 Mg + O 2 2 MgO 2 moles : 1 mole : 2 moles

Steps to Solve a Stoichiometry Problem Make sure the equation is BALANCED. Label the QUANTITIES GIVEN. DRAW a DIMENSIONAL ANALYSIS FRAME: Grams starting X FW starting, g 1 mole starting Moles starting Moles ending 1 mole ending FW ending, g

Fill in the frame with the MOLES and FWs as indicated on the frame. Complete the math across the top of the frame, then across the bottom. Finally divide the TOP by BOTTOM and ROUND to SIGNIFICANT FIGURES.

Practice Problem If 75.2 g of KClO 3 decomposes, how many grams of O 2 will be produced? 2 KClO 3 2 KCl + 3 O g ? g ✔ balanced

75.2g KClO 3 = FW KClO 3 X X g KClO 3 1 mole KClO 3 1 mole KClO 3 2 mole KClO 3 2 mole KClO 3 3 mole O 2 3 mole O 2 1 mole O 2 1 mole O g O g O 2 FW O 2 Mole Ratio Mole Ratio ______ g O 2

75.2g KClO 3 = X X g KClO 3 1 mole KClO 3 1 mole KClO 3 2 mole KClO 3 2 mole KClO 3 3 mole O 2 3 mole O 2 1 mole O 2 1 mole O g O g O g O 2 ____________________ g O 2 ____________________ 245.1

= g ≈ 29.5 g O 2 = g ≈ 29.5 g O 2

MORE PRACTICE #1) If 75.2 grams of Calcium Sulfate are reacted with an unlimited amount of Sodium Phosphate, what mass of Calcium Phosphate will be formed according to the reaction: CaSO 4 + Na 3 PO 4  Ca 3 (PO 4 ) 2 + Na 2 SO 4

#2) How many grams of Potassium are needed to form grams of Potassium Carbonate according to the reaction below? K + Li 2 CO 3  K 2 CO 3 + Li