CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (g) 1 mol2 mol1 mol2 mol Stoichiometry of the reaction FIXED ratio for each reaction If the amount of one chemical is known we can calculate how much other chemicals are required or produced.
4 NH 3 (g) + 5 O 2 (g) → 4 NO (g) + 6 H 2 O (g) 6.02 mol of NH 3 is used in the above reaction. How many moles of O 2 is required to react with all the NH 3 ? How many moles of H 2 O will be produced?
2NH 3 (g) + 3CuO(s) → N 2 (g) + 3Cu(s) + 3H 2 O(g) 2 mol 3 mol1 mol3 mol 6.04 g xy 6.04 g ÷ (14.01 g/mol x 1 + 1.008 g/mol x 3) = 0.355 mol 0.355 mol mass? Mass of CuO = 0.533 mol x 79.55 g/mol = 42.4 g Mass of N 2 = 0.178 mol x 28.02 g/mol = 4.99 g
CH 4 + 2O 2 → CO 2 + 2H 2 O 0 mol 1 mol2 mol initial: 0 mol 1 mol 2 mol final: 1 mol 0 mol initial: ? mol final: The actual amount of reactants consumed and actual amount of products produced agree with the stoichiometry.
CH 4 + 2O 2 → CO 2 + 2H 2 O 1 mol 0 mol initial: 1 mol CH 4 requires 2 mol O 2, available O 2 is 1 mol: limiting reagent. (1 − 0.5) mol 0 mol = 0.5 mol= 1 mol Result: 1 mol O 2 will be consumed completely. = 0.5 mol consumed: 1 mol x x = 0.5 mol final: yz CH 4 will have leftover: excess reagent. y = 0.5 mol
The reactant of which there are fewer moles than the stoichiometry requires is the limiting reagent. The reactant of which there are more moles than the stoichiometry requires is the excess reagent. Chemical reactions always occur according to the stoichiometry, therefore the limiting reagent is consumed and the excess reagent has leftover. The amount of products is determined by the amounts of reagents that are actually consumed.
2 slices of bread + 1 slice if ham → 1 sandwich 4 slices of bread + 1 slice if ham → excess reagent leftover limiting reagentamount of product 1 sandwich + 2 slices of bread
For the following reaction, if a sample containing 18.1 g of NH 3 reacted with 90.4 g of CuO, which is the limiting reagent? How many grams of N 2 will be formed? How many grams of excess reagent will be left over? If 6.63 g of N 2 is actually produced, what is the percent yield? NH 3 (g) + CuO(s) → N 2 (g) + Cu(s) + H 2 O(g)
1)Make sure the equation is balanced. 2)Find the moles of each reactant: moles = mass in gram / molar mass 3)Pick up any reactant, say A, and use the stoichiometry to calculate the required amount of the other reactant B. 4)Compare the required amount of B with the available amount of B. a) If required > available, then B is the limiting reagent and A is the excess reagent. b) If required < available, then B is the excess reagent and A is the limiting reagent. 5)Use the amount of the limiting reagent and the stoichiometry to calculate the amount of any product and the amount of the excess reagent that has been consumed. 6)Leftover excess reagent = available − consumed 7)If actual yield is given percent yield = (actually yield / theoretical yield) x 100% Procedure for limiting/excess reagent calculations aA + bB → cC + dD
salts (ionic compounds): NaCl, K 2 SO 4, … strong acids: HCl, HNO 3, H 2 SO 4, HClO 4 strong bases: NaOH, KOH Bases: compounds that give OH − when dissolved in water. Strong electrolytes weak acids: acetic acid: HC 2 H 3 O 2 weak bases: ammonia: NH 3 Weak electrolytes
Example 4.5, page 153 25.5 g of KBr is dissolved in water and forms a solution of 1.75 L. What is the molarity of the solution? Example 4.6, page 154 How many liters of a 0.125 mol/L NaOH solution contains 0.255 mol of NaOH?
How to prepare 1.00 L of NaCl aqueous solution with a molarity of 1.00 mol/L? 1.00 mol NaCl + 1.00 L of H 2 O = 1.00 mol/L NaCl (aq)
Solution Dilution Concentrated solutions for storage, called stock solutions stock solution + water desired solution
moles of solute before dilution = moles of solute after dilution M 1 V 1 = M 2 V 2 M 1 : molarity of concentrated solution V 1 : volume of concentrated solution M 2 : molarity of diluted solution V 2 : volume of diluted solution Example on page 155 A lab procedure calls for 3.00 L of a 0.500 mol/L CaCl 2 solution. How should we prepare it from a 10.0 mol/L stock solution?
Read acid-base titration starting on page 168. Read the online instruction for the titration experiment. End point: light pink
The titration of a 25.00-mL sample of an HCl solution of unknown concentration requires 32.54 mL of a 0.100 mol/L NaOH solution to reach the equivalence point. What is the concentration of the unknown HCl solution in mol/L? M NaOH V NaOH = M HCl V HCl HCl(aq) + NaOH(aq) → H 2 O(l) + NaCl(aq)
Types of reactions Precipitation reactions Acid-Base reactions Oxidation-Reduction reactions
Reactions that involve electron transfer are called oxidation-reduction reactions, or redox reactions. 2Mg(s) + O 2 (g) → 2MgO(s)
Oxidation number (state) 1) For atoms in its elemental form, oxidation number = 0 A way to keep track of the electrons gained or lost Na, Ar, O 2, N 2, O 3, P 4, S 8 2) For monatomic ion, oxidation number = charge of the ion Na +, Ca 2+, Co 2+, Co 3+, Cl −, O 2− NaCl, Na 2 O, CaCl 2, CaO, CoCl 2, CoCl 3, Co 2 O 3, CoO
3) In covalent compounds Remember O: −2H: +1F: −1 In a neutral compound, the sum of the oxidation number = 0 In a polyatomic ion, the sum of the oxidation number = ion charge CO, CO 2, SF 6, SF 4, H 2 S, NH 3, P 2 O 5, N 2 O 3 NO 3 −, SO 4 2−, NH 4 +, Cr 2 O 7 2−, MnO 4 −
Reactions that cause change of oxidation numbers are called redox reactions. Element loses electrons → its oxidation number increases → element is oxidized → oxidation reaction Substance that contains the oxidized element is call the reducing agent. Substance that contains the reduced element is call the oxidizing agent. Element gains electrons → its oxidation number decreases → element is reduced → reduction reaction