Usual Atomic Charges of Main Group Elements

Transcription

1 Usual Atomic Charges of Main Group Elements Examples SO 3 sulfur trioxide CO 2 carbon dioxide Al 2 O 3 aluminum trioxide IF 7 iodine heptafluoride Fig. 2-6, p.63

2 Chemical Equations A chemical equation just like a mathematical equation is a way to express, in symbolic form, the reactions occurring in a chemical system. Balancing chemical equations Reaction stoichiometry Reagents limiting the extent of reaction Acid-base reactions Oxidation states of reactants and products

3 Types of Chemical Reactions Combination Reactions: Atoms or molecules combine to form a new molecule 2H 2 + O 2 2H 2 O ClO + NO 2 ClNO 3 ClO + ClO ClOOCl

4 Types of Chemical Reactions Decomposition Reactions: A molecule breaks apart to form different molecules or atoms NO 2 + sunlight O + NO 2C 7 H 5 (NO 2 ) 3 + heat 7CO(g) + 7C + 5H 2 O(g) + 3N 2 (g)

5 Types of Chemical Reactions Displacement Reactions: An atom or molecule displaces an atom or molecule in the reaction partner H + Cl 2 HCl + Cl 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g)

6 Types of Chemical Reactions Exchange Reactions: The components of two compounds are exchanged BaCl 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaCl(aq) 3NaOH(aq) + H 3 PO 4 (aq) Na 3 PO 4 (aq) + 3H 2 O(l)

7 Balancing Chemical Equations Remember Conservation of Mass rule-- that matter is neither created nor destroyed in a chemical reaction. A balanced chemical equation must have the same number of each type of atom on the reactant side as on the product side.

8 Balancing Chemical Equations Process for balancing chemical equations: 1. Determine correct chemical formulas of all reactants and products 2. Start with heavier atoms--balance number of these on reactant and product sides of equation 3. If elements appear in equation as either reactants or products, balance these last 4. Electrical charge must be balanced

9 Balanced Chemical Equations Examples Write a balanced chemical equation for the reaction of Fe(III) with oxygen to form iron oxide 1. Fe + O 2 Fe 2 O Fe + O 2 Fe 2 O 3 Fe is currently balanced 3. 2 Fe + 3/2 O 2 Fe 2 O 3 O is now balanced coefficients must usually be integer numbers-- multiply everything by 2 to remove 3/2 denominator: 4 Fe + 3 O 2 2 Fe 2 O 3

10 Balanced Chemical Equations Examples Write a balanced chemical equation for the combustion of methane combustion is reaction with oxygen producing a flame. Complete combustion produces only CO 2 and H 2 O.

11 Balanced Chemical Equations Examples 1. CH 4 + O 2 CO 2 + H 2 O 2a. CH 4 + O 2 CO 2 + H 2 O C is now balanced 2b. CH 4 + O 2 CO H 2 O H is now balanced 3. CH O 2 CO H 2 O CH O 2 CO H 2 O O is now balanced

12 Balanced Chemical Equations Examples Combustion of propane, C 3 H 8 1. C 3 H 8 + O 2 CO 2 + H 2 O 2a. C 3 H 8 + O 2 3 CO 2 + H 2 O 2b. C 3 H 8 + O 2 3 CO H 2 O 3. C 3 H O 2 3 CO H 2 O 4. No charges to balance C 3 H O 2 3 CO H 2 O

13 Balanced Chemical Equations Examples Reaction of ammonium perchlorate with aluminum NH 4 ClO 4 (s) + Al(s) Al 2 O 3 (s) + N 2 (g) + HCl(g) + H 2 O(g) 2a. Balance N: 2 NH 4 ClO 4 (s) + Al(s) Al 2 O 3 (s) + N 2 (g) + HCl(g) + H 2 O(g) 2b. Balance Cl: 2 NH 4 ClO 4 (s) + Al(s) Al 2 O 3 (s) + N 2 (g) + 2 HCl(g) + H 2 O(g)

14 Balanced Chemical Equations Examples Reaction of ammonium perchlorate with aluminum (con t.) 2c. Balance H: 2 NH 4 ClO 4 (s) + Al(s) Al 2 O 3 (s) + N 2 (g) + 2 HCl(g) + 3 H 2 O(g) 2d. Balance O: 2 NH 4 ClO 4 (s) + Al(s) 5/3 Al 2 O 3 (s) + N 2 (g) + 2 HCl(g) + 3 H 2 O(g)

15 Balanced Chemical Equations Examples Reaction of ammonium perchlorate with aluminum (con t.) 3a. Balance Al: 2 NH 4 ClO 4 (s) + 10/3 Al(s) 5/3 Al 2 O 3 (s) + N 2 (g) + 2 HCl(g) + 3 H 2 O(g) 3b. Remove fractional coefficients (multiply by 3): 6 NH 4 ClO 4 (s) + 10 Al(s) 5 Al 2 O 3 (s) + 3 N 2 (g) + 6 HCl(g) + 9 H 2 O(g)

16 Balanced Chemical Equations Examples Mg(s) + H 3 O + (aq) Mg 2+ (aq) + H 2 (g) + H 2 O( ) 1. As written, the only element out of balance is the hydrogen balance H first: Mg(s) + 2 H 3 O + (aq) Mg 2+ (aq) + H 2 (g) + 2 H 2 O( ) 2. Check charge balance: left-hand side has total charge of +2 (2 * 1+ charge on hydronium ion) Right-hand side has total charge of +2 (2+ charge on Mg ion)

17 Stoichiometry of Chemical Reactions For the generic chemical reaction aa + bb + xx + yy + A, B, X, and Y represent the atoms or molecules reacting and forming, and the coefficients a, b, x, and y represent the stoichiometric coefficients they tell how many moles of one substance reacts with another substance to form some number of moles of the products.

18 Stoichiometry It is important to remember that the stoichiometric coefficients represent numbers of moles, not masses of reactants and products

19 Stoichiometry How much carbon dioxide is produced by burning one gallon of gasoline? Gasoline is composed of many hydrocarbons, but let s assume they are all iso-octane (C 8 H 18 ) Isooctane has a density of g ml -1 1 gal = L = 3785 ml

20 Stoichiometry Mass of isooctane ( g ml -1 ) (3785 ml) = 2642 g C 8 H 18 Molar mass: 8( g mol -1 ) + 18( g mol -1 ) Moles C 8 H 18 : = g mol -1 (2642 g) / ( g mol -1 ) = mol C 8 H 18

21 Stoichiometry Balanced chemical equation: 2 C 8 H O 2 16 CO H 2 O Determine moles of CO 2 produced: (23.13 mol C 8 H 18 ) (16 mol CO 2) (2 mol C 8 H 18 ) = mol CO 2

22 Stoichiometry Determine mass of CO 2 : (185.0 mol CO 2 ) ( g mol -1 ) = 8142 g CO 2 = kg CO 2 per gallon of gasoline consumed How much CO 2 is emitted by CA every year? Californians consume ~1.4 x gallons of gasoline each year. (1.4 x gal) (8.142 kg CO 2 gal -1 ) = 1.1 x kg CO 2

23 Stoichiometry The following general expression applies to stoichiometry problems: Moles B = (Coefficient B) (Coefficient A) (Moles A)

24 Limiting Reagents Suppose the amount of one the reactant in a chemical reaction is insufficient to allow the reaction to proceed to completion. That reactant is called the limiting reagent. The limiting reagent limits how much product can formed in a reaction.

25 Limiting Reagents 6 NH 4 ClO 4 (s) + 10 Al(s) 5 Al 2 O 3 (s) + 3 N 2 (g) + 6 HCl(g) + 9 H 2 O(g) If we begin with 1.00 kg ammonium perchlorate and kg aluminum, how many moles of gaseous product will be produced? 1. Which reactant will be consumed first? (1000 g NH 4 ClO 4 )/( g mol -1 ) = 8.51 mol NH 4 ClO 4 (100 g Al)/( g mol -1 ) = 3.71 mol Al

26 Limiting Reagents 1a. How much Al required for NH 4 ClO 4 to react completely? (8.51 mol NH 4 ClO 4 ) (10 mol Al) (6 mol NH 4 ClO 4 ) = 14.2 mol Al Since we only have 3.71 mol Al, there is insufficient Al for the NH 4 ClO 4 to react completely, so Al is the limiting reagent. 2. Determine moles of each gaseous product formed in reaction:

27 Limiting Reagents (3.71 mol Al) (3 mol N 2) (10 mol Al) = 1.11 mol N 2 (3.71 mol Al) (6 mol HCl) (10 mol Al) = 2.23 mol HCl (3.71 mol Al) (9 mol H 2O) (10 mol Al) = 3.34 mol H 2 O Total gas phase product = 6.68 mol

28 Limiting Reagents Example Ammonia reacts with nitric oxide to form nitrogen gas and water: NH 3 + NO N 2 + H 2 O If 71.4 g NH 3 reacts with g NO, how much N 2 and H 2 O will be produced? Step 1: Balance the chemical equation

29 Limiting Reagents Example (con t.): NH 3 + NO N 2 + H 2 O Step 1: Balance the chemical equation 2 NH 3 + NO N H 2 O H balanced 2 NH NO N H 2 O O balanced 2 NH NO 5/2 N H 2 O N balanced 4 NH NO 5 N H 2 O remove fractional coefficient

30 Limiting Reagents Example (con t.): NH 3 + NO N 2 + H 2 O Step 2: Determine moles of reactants 71.4 g NH3 = 5.09 mol NH g / mol g NO g/mol = 5.62 mol NO Step 3: Determine which is limiting reagent (4 mol NH ) 5.62 mol NO) (6 mol NO) mol NH 3 ( = NO is limiting reagent

31 Limiting Reagents Example (con t.): NH 3 + NO N 2 + H 2 O Step 4: Determine amount of products ( 5 mol N ) (5.62 mol NO) 2 (6 mol NO) (4.68 mol (6 mol H O) (5.62 mol NO) 2 (6 mol NO) (5.62 mol N H 2 2 = 4.68 mol ) ( g/mol) = 131g N = 5.62 mol N H O O) ( g/mol) = 101g H O