Name: Hr: Understand and be able to explain all of the key concepts. Define and understand all of the survival words Memorize the names and symbols for these elements: (Ag, Al, Ar, As, Au, B, Ba, Be, Br, C, Ca, Cd, Cs, Cl, Co Cr, Cu, F, Fe, Fr, H, He, Hg, I, K, Kr, Li, Mg, Mn, N, Na, Ne, Ni, O, P, Pb, Rn, S, Sc, Si, Sn, Sr, Ti, U, Xe, Zn) Know the charges (oxidations numbers) of elements found in the s & p blocks. Know which elements are transition metals and therefore require the charge in parenthesis when being named. [ex. Sn (II) or Co (III) ] Know the charges of Ag +1, Zn +2, Cd +2, Al +3, and Ga +3 (exceptions to the transition area) Know how to use the Activity Series of Metals and Halogens for single replacement reactions Know how to use the Solubility Rules Know how to predict the products of any reaction Know how to balance a chemical reaction Know how to determine the net ionic equation. Review all classwork (including your lab notebook) and quizzes. Key Concepts Hydrocarbons Because carbon has four valence electrons, carbon atoms always form four covalent bonds. Molecules of hydrocarbons are nonpolar Carbon atoms can form chains that are named using the following naming convention: o Prefix is determined by the number of carbons 1 = meth 5 = pent 2 = eth 6 = hex 3 = prop 7 = hept 4 = but 8 = oct o Suffix is determined by the type of bond Alkane C n H 2n+2 (all bonds are single) Alkene C n H 2n (one bond is a double) Alkyne C n H 2n-2 (one bond is a triple) 9 = non 10 = dec 12.1 The Arithmetic of Equations A balanced chemical equation provides the same kind of quantitative information that a recipe does. Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in the reaction. A balanced chemical equation can be interpreted in terms of different quantities including numbers of atoms, molecules, or moles; mass; and volume. Mass and atoms are conserved in every chemical reaction. 1 P a g e
12.2 Chemical Calculations In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product, or between moles of products. In a typical stoichiometric problem, the given quantity (starting quantity) is first converted to moles. Then the mole ratio from the balanced equation is used to calculate the moles of the wanted substance. Finally, the moles are converted to any other unit of measurement related to the unit mole. 12.3 Limiting Reagent and Percent Yield In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. The percent yield is a measure of the efficiency of a reaction performed in the laboratory. Survival Words Actual yield (372) Atom (101) Balanced equation (325) Chemical equation (353) Coefficients (325) Excess reagent (369) Limiting reagent (369) Molar mass (294) Key Equations Percent Yield = Mole (290) Mole ratio (359) Molecule (214) Percent yield (372) Representative Particle (290) Stoichiometry (354) Theoretical yield (372) Yield (323) Actual Yield Theoretical Yield 100 NOTE: You are responsible for memorizing the polyatomic ion list and the diatomic elements! 2 P a g e
Review Questions You must show all work for credit! You will need a separate piece of paper. 1. What changes during a chemical reaction? 2. What remains the same during a chemical reaction? 3. Balance the following equations a. _1 Ba + _1 H 2 SO 3 _1 BaSO 3 + _1 H 2 b. _1 C 3 H 8 + _5 O 2 3_CO 2 + 4_H 2 O c. _1 Sn(NO 2 ) 2 + _2 KI _1 SnI 2 + _2 KNO 2 d. _3 Ni + _2 O 2 _1 Ni 3 O 4 e. _2 H 2 + _1 O 2 _2 H 2 O f. _2 C 7 H 14 + _14 O 2 _7 CO 2 + _14 H 2 O 4. 4 Al + 3 O 2 2 Al 2 O 3 a. Draw a particle diagram for the equation shown. =Al =O b. How many total molecules are on the reactant side? 3 (plus 4 individual atoms) How many atoms? 10 c. How many total molecules are on the product side? 2 How many atoms? 10 d. How many different compounds exist in the reaction? 2 compounds plus 1 element 5. Mg 3 N 2 + 3 H 2 O 3 MgO + 2 NH 3 =Mg =H =N =O a. Draw a particle diagram for the equation shown. b. How many total molecules are on the reactant side? 4 How many atoms? 14 c. How many total molecules are on the product side? 5 How many atoms? 14 d. How many different compounds exist in the reaction? 4 e. If the mass of products is 34 grams, what was the mass of the reactants? 34 grams 3 P a g e
6. Interpret the given equation in terms of relative numbers of a) representative particles, b) numbers of moles, and c) masses of reactants and products. 2K (s) + 2H 2 O (l) 2KOH (aq) +H 2 (g) 2K +2 H 2 O 2 KOH + H 2 2 atoms 2 molecules 2 formula units (ionic) 1 molecule 2 moles 2 moles 2 moles 1 mole 2(39.1) = 78.2 g 2 (2(1) + 16) = 32 g 2 (39.1+16+1) = 112.2 g 2(1) = 2 g 7. Balance the equation: C 2 H 5 OH (l) + O 2 (g) CO 2 (g) + H 2 O (g). Show the balanced equation obeys the law of conservation of mass by calculating the masses of reactants and products. C 2 H 5 OH + 3 O 2 2 CO 2 + 3 H 2 O Reactants: 2(12) + 5(1) + 16 + 1 + 3(2(16)) = 142 g Products: 2 (12 + 2(16)) + 3 (2(1) + 16) = 142 g 8. Acetylene gas (C 2 H 2 ) and Calcium hydroxide are produced by adding water to calcium carbide (CaC 2 ). a) Write the balanced chemical equation. b) How many grams of acetylene are produced by adding excess water to 5.00 grams of calcium carbide? a) CaC 2 + 2 H 2 O C 2 H 2 + Ca(OH) 2 b) 5.00 g CaC 2 / (40.1 + 2(12)) = 0.078 mol E CaC 2 + 2 H 2 O C 2 H 2 Ca(OH) 2 B 0.078 XS 0 0 C -0.078-0.156 +0.078 +0.078 A 0 XS 0.078 mol acetylene x (2(12) + 2(1)) = 2.03 g 0.078 (these are really the only columns you need to do for this problem) 0.078 9. Using the equation you balanced in the problem above, determine how many moles of water are needed to react completely with 5.00 grams of calcium carbide. (work in table above) 0.156 mol H 2 O x (2(1) + 16) = 2.81 g 10. The last step in the production of nitric acid is the reaction of nitrogen dioxide with water. 3NO 2 (g) + H 2 O (l) 2HNO 3 (aq) + NO (g) How many grams of nitrogen dioxide must react with water to produce 5.00 x 10 22 molecules of nitrogen monoxide? 5.00 x 10 22 molecules / 6.02 x 10 23 = 0.083 mol NO 4 P a g e
E 3 NO 2 + H 2 O 2 HNO 3 + NO B 0.25 0 C (0.083 * 3 =) 0.25 +0.083 A 0 0.083 0.25 mol NO 2 x (14 + 2(16)) = 11.5 g NO 2 11. How are mole ratios used in chemical calculations? Mole ratios are used to compare the amount of each reactant and product to each other. 12. Write a sequential list of steps required to solve most typical stoichiometric problems. Using a BCA table: 1. Start with a balanced equation. 2. Convert any values given into moles (if needed). 3. Complete the BCA table using the coefficients of the balanced equation to convert one change value into another. 4. Convert the mol value at the end into another unit (if needed). Using a molar road: 1. Start by writing the value you are given. 2. Convert the value given into moles (if needed). 3. Convert into moles of the other substance with the mole to mole ratio. 4. Convert from moles into a new unit (if needed). 13. Write the 12 mole ratios that can be derived from the equation for the combustion of isopropyl alcohol. 2C 3 H 7 OH (l) + 9O 2 (g) 6CO 2 (g) + 8H 2 O (g) 2C 3 H 7 OH: 9O 2 2C 3 H 7 OH: 6CO 2 2C 3 H 7 OH: 8H 2 O 9O 2 : 6CO 2 9O 2 : 8H 2 O 6CO 2 : 8H 2 O **And reciprocals** 14. a) Write the equation for the complete combustion of ethene. C 2 H 4 + 3 O 2 2 CO 2 + 2 H 2 O b) If 2.7 mol of ethane is reacted with 6.3 mole of oxygen gas, identify the limiting reagent. E C 2 H 4 + 3 O 2 2 CO 2 2 H 2 O B 2.7 6.3 0 0 C -2.1-6.3 +4.2 +4.2 5 P a g e
A 0.6 0 4.2 4.2 1mol C 2 H 4 = 2.7 3mol O 2 x x = 8.1 mol O 2, so if all 2.7 mol of C 2 H 4 react, then we need 8.1 mol of O 2. We are only given 6.3 mol, so O 2 is the limiting reactant. c) Calculate the number of moles of water produced. 4.2 mol 15. When 84.8 grams of iron (III) oxide reacts with an excess of carbon monoxide, iron and carbon dioxide are produced. a) Write the balanced chemical equation. Fe 2 O 3 + 3 CO 2 Fe + 3 CO 2 b) What is the theoretical yield of this reaction for each product? 84.8 g Fe 2 O 3 / (2(55.8)+3(16)) = 0.53 mol E Fe 2 O 3 + 3 CO 2 Fe + 3 CO 2 B 0.53 XS 0 0 C -0.53-1.59 +1.06 +1.59 A 0 XS 1.06 mol 1.59 mol 16. If 50.0 grams of silicon dioxide is heated with an excess of carbon, 27.9 grams of silicon carbide is produced. SiO 2 (s) + 3C (s) SiC (s) + 2CO (g) What is the percent yield of this reaction? 50g SiO 2 / (28.1 + 2(16)) = 0.83 mol E SiO 2 + 3C SiC + 2CO B 0.83 0 C -0.83 +0.83 A 0 0.83 0.83 mol SiC x (28.1 + 12) = 33.28 g % yield = 27.9 g / 33.28 g = 83.8 % 6 P a g e
17. If you were given 12 grams of Mg and excess amounts of 6 molar hydrochloric acid (6 mol HCl/L H 2 O), exactly how much hydrochloric acid (in ml) would you need to react all of the magnesium? What is the balanced chemical equation for this chemical reaction? Balanced chemical equation: Mg + 2 HCl H 2 + MgCl 2 12 g Mg / 24.3 g = 0.49 mol E Mg + 2 HCl H 2 + MgCl 2 B 0.49 XS C -0.49-0.98 A 0 XS 0.98 mol HCl x L = 6 mol HCl 1 L X = 0.163 L HCl ( = 163 ml) 7 P a g e