Reactions Conservation of Matter, Stoichiometry, and Moles

Standards 3. The conservation of atoms in chemical reactions leads to the principle of conservation of matter and the ability to calculate the mass of products and reactants. As a basis for understanding this concept: a. Students know how to describe chemical reactions by writing balanced equations. 3. b. Students know the quantity one mole is set by defining one mole of carbon-12 atoms to have a mass of exactly 12 grams. 3. c. Students know one mole equals 6.02 × 10^23 particles (atoms or molecules). 3. d. Students know how to determine the molar mass of a molecule from its chemical formula and a table of atomic masses and how to convert the mass of a molecular substance to moles, number of particles, or volume of gas at standard temperature and pressure. 3. e. Students know how to calculate the masses of reactants and products in a chemical reaction from the mass of one of the reactants or products and the relevant atomic masses. 3. f.* Students know how to calculate percent yield in a chemical reaction. 3. g.* Students know how to identify reactions that involve oxidation and reduction and how to balance oxidation-reduction reactions.

Example Reaction Reactants  Productswhat you start out withend up with

Reaction Symbols SymbolMeaning (s), (l), (g) Substance is a solid, liquid, or gas (aq)Aqueous, substance is dissolved in H 2 O “Produces” or “makes” “Produces” through reversible reaction heat or Δ Heat is added to the reactants Pt A catalyst is used to speed up the reaction

Conservation of Mass and Atoms The total mass of the reactants is the same as the total mass of the products. It doesn’t change. The number of each type of atom for the reactants is the same as number of each type of atom for the products. It doesn’t change.

Example Reaction #1 2 H 2 + O 2  2 H 2 O hydrogen oxygen water gas gas +

Example Reaction #1 2 H 2 + O 2  2 H 2 O hydrogen oxygen water gas gas grams32.00 grams grams

Basic Types of Reactions CombinationSingle Substitution (aka Synthesis)(aka Displacement) A + B  AB A + BC  B + AC DecompositionDouble Substitution AB  A + B AB + CD  CB + AD

Example Reaction #2 2 NH 3 (l)  N 2 (g) + 3 H 2 (g) liquid nitrogen hydrogen ammonia gas gas +

Example Reaction # 2 H H H H + 2 NH 3 (l)  N 2 (g) + 3 H 2 (g) liquid nitrogen hydrogen ammonia gas gas H H N H H H N H H H N N

Combustion of Methane CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) methaneoxygen carbonwater dioxide + +

Combustion Reactions Complete Combustion: Fuel + O 2  CO 2 + H 2 O Incomplete Combustion: Fuel + O 2  CO + H 2 O Example Fuels: Methane (CH 4 ), Propane (C 3 H 8 ), Octane (C 8 H 18 ), Wood, etc.

? ? ? How much?

1 cup 1.5 cups 6 ounces 3 eggs = 26 cookies

How much? CH 4 (g) + H 2 O (g)  CO (g) + 3 H 2 (g) Methanewater carbon hydrogen monoxide 6.06 grams grams + +

Stoichiometry Stoichiometric calculations use mole ratios from the balanced chemical equations to predict amounts of products or reactants.

What is the mole ratio between H 2 and O 2 ? 2 H 2 + O 2  2 H 2 O hydrogen oxygen water gas gas +

2 H 2 + O 2  2 H 2 O hydrogen oxygen water gas gas What is the mole ratio between H 2 and O 2 ? 2 mol H 2 1 mol O 2

2 H 2 + O 2  2 H 2 O hydrogen oxygen water gas gas What is the mole ratio between H 2 O and H 2 ? 2 mol H 2 O 2 mol H 2 1 mol H 2 O 1 mol H 2 =

Stoichiometric Calculation #1 How many moles of oxygen gas (O 2 ) are needed to react completely with 6.0 moles of hydrogen gas (H 2 )? 1 mol O 2 2 mol H mol H 2 1 × = 3.0 mol O 2 = mol O 2 2 H 2 + O 2  2 H 2 O

Stoichiometric Calculation #2 What is the mass of ammonia (NH 3 ) needed to produce 9.00 moles of hydrogen gas (H 2 )? 2 mol NH 3 3 mol H mol H 2 1 × = 102 g NH 3 = 9.00 × 2 × g NH 3 2 NH 3 (l)  N 2 (g) + 3 H 2 (g) g NH 3 1 mol NH 3 × =

Stoichiometric Calculation #3 How many liters of carbon dioxide (CO 2 ) are produced from the complete combustion of 2.00 moles of propane (C 3 H 8 )? 3 mol CO 2 1 mol C 3 H mol C 3 H 8 1 × = 134 L CO 2 = 2.00 × 3 × L CO 2 C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) 22.4 L CO 2 1 mol CO 2 × =

The Breakfast of Chemists!  16 products reactants

Limiting and Excess Reactants Limiting reactant – the reactant that you will run out of. Excess reactant – the reactant that you have extra amounts of.

Steps for Limiting Reactant Problems 1.Divide the reactant moles by their coefficients. Compare the results. 2. The lower one is the limiting reactant. The higher one is the excess reactant. 3.Only use the limiting reactant for stoichiometric calculations.

Limiting Reactant Problem #1 If you have 3 moles of magnesium (Mg) and 4 moles of hydrochloric acid (HCl), which substance is the limiting reactant? Which substance is the excess reactant? Mg (s) + 2 HCl (aq)  MgCl 2 (aq) + H 2 (g) 3 mol4 mol 3 mol 1 = 3 4 mol 2 = 2 limitingexcess >

Limiting Reactant Problem #2 How many moles of magnesium chloride (MgCl 2 ) will be produced? 1 mol MgCl 2 2 mol HCl 4.0 mol HCl 1 × = 2.0 mol MgCl 2 = mol MgCl 2 Mg (s) + 2 HCl (aq)  MgCl 2 (aq) + H 2 (g) 3 mol4 mol

Add Breakfast of Chemists Art and intro Limiting Reactant (Reagent) Practice

Cookie recipe – want to know precise amounts – so it doesn’t turn out wrong, & so we don’t waste ingredients Learn how using only the periodic table & the chemical reaction to have the precise amount What is a mole, molar mass, # of particles Balanced chemical equations

Combustion Reactions Combustion is a double substitution reaction. For a single reaction, there are 2 reactants: the fuel (ex. CH 4 ) and oxygen (O 2 ). With complete combustion there are 2 products: carbon dioxide (CO 2 ) and water (H 2 O). Incomplete combustion occurs when there is not enough oxygen (O 2 ) present. For this the 2 products are: carbon monoxide (CO) and water (H 2 O).

Ne F ONC B Be He Li H KrAr Cl Br Xe ISPSi Mg Al Ca Na K

Ne F ONC B Be He Li H KrAr Cl Br Xe ISPSi Mg Al Ca Na K

4 e – in valence shell