Balanced Chemical Equations, Stoichiometry & Atom Dimensional Analysis Problems 1. Change a speed of 72.4 miles per hour to its equivalent in meters per second (1miles = 1.609km). 32.35 m/s 2. The density of mercury is 13.6 g/mL. What is the mass in kilograms of a 2 L commercial flask of mercury? 27.2kg Balanced Chemical Equations & Stoichiometry • Follow the law of conservation! If you use up something on the reactant side then there must be the same amount of resulting product. Problems 1) S 8 + 8 O 2 → 8 SO 2 2) 2 NH 3 + 3 CuO → 3 Cu + N 2 + 3 H 2 O 3) P 4 O 10 + 6 H 2 O → 4 H 3 PO 4 4) Zn + 2 AgCl → ZnCl 2 + 2 Ag 5) 2 HCl + CaCO 3 → CaCl 2 + H 2 O + CO 2 Atoms • Protons are positively charged particles that are found in the nucleus • Neutrons are non-charged particles also found in the nucleus. Atoms with the same number of protons but different number of neutrons are isotopes. • Electrons are negatively charged particles that surround the nucleus. They are responsible for a lot of properties exhibited by atoms. Problems An atom has 25 protons and a molar mass of 55, how many neutrons and electron does it have? What if the molar mass was 57, would the number of neutrons and electrons change?
Neutrons= 30, Electrons= 25. Neutrons= 32, Electrons 25 Reactions Limiting Reactants • A reactant that determines how much product is produced. • In order to determine the limiting reactant, compare the yields of the reactants, whichever one makes the least amount of product will be the limiting reactant. Empirical & Molecular Formula • The simplest whole number ratio of moles of each element present in a compound. • You would determine the empirical formula by solving for the number of moles of each element. Once those are determined, divide the smallest number of moles and attain ratios. If the ratios are not integers multiply them all by a common value till you attain integers. • Molecular formula is based on the exact number of moles of an element in a compound. • Ex: Glucose has a molecular formula of C 6 H 12 O 6 but its empirical formula is CH 2 O Combustion Reactions • Combustion reactions are exothermic reactions that occur between a fuel and an oxidant. For our purposes, the fuel will be an organic compound (hydrocarbons) and the oxidant would be O 2 . They are exothermic but however, In order for combustion reactions to occur there must be an input of energy. Problems The combustion of 20.63g of Compound S, containing only C,H, and O, with excess oxygen gives 57.94g of CO 2 and 11.85g of H 2 O. (a) Calculate how many grams of C,H, and O were present in the original sample of S. (b) Determine the empirical formula of S. (c) If the molar mass of compound S is 282 g/mol, what would be its molecular formula? a) C=15.8g, H=1.3g, O=3.53g b) C 6 H 6 O c) C 18 H 18 O 3 Reactions (cont)
Acid/Base Reactions Nuetralization Moles of acid = moles of base Solute what is being dissolved Solvent what you are dissolving it in Solution solute + solvent Molarity = moles of solute Molality = moles of solute X = moles of one species Liters of solution kg of solvent total amount of moles Problems How many liters of .75 M KOH are required to completely neutralize a 75mL of a .50M solution
You've reached the end of your free preview.
Want to read all 17 pages?