1. a review of compound and reaction stoichiometry, including solutions and gases

2. Compound Stoichiometry Deals with the mole relationship between elements in a compound, so convert information given in other units (grams, liters, etc.) to moles before proceeding. Empirical formula: formula showing the simplest possible whole number ratio of elements in a compound Molecular formula: formula showing the actual number of atoms of each element in one molecule of a compound (or moles of each element in one mole of the compound); always a whole number multiple of the empirical formula

3. Reaction Stoichiometry Deals with the mole relationship between reactants and products in a chemical reaction, so convert information given in other units (grams, liters, etc.) to moles. (If asked for another unit, convert to the desired unit at the end.) Required a balanced chemical equation—if one is not given, write it yourself! Limiting reactant: reactant that runs out and stops the reaction (If you are given quantity information about more than one reactant, assume you must identify this!) Excess reactant(s): any reactant(s) other than the limiting reactant; some will be leftover after the reaction stops

4. Solution and Gas Stoichiometry Molarity (M) = mol solute / L solution PV = nRT P in atm V in L n in mol R = 0.0821 L·atm/mol·K T in K Standard molar volume: 1 mol = 22.4L of gas only at STP

5. MC Question 1 Molar mass of CO 2 = 44g/mol 88/44 = 2 mol CO 2 = 2 mol C from original compound Molar mass of H 2 O = 18g/mol 27/18 = 1.5 mol H 2 O = 3 mol H from original compound The only answer choice with a 2:3 ratio of C to H is D

6. MC Question 2 Water is a little more than half the hydrate’s total mass (54.3%). The dry Na 2 CO 3 has a molar mass of 106 g/mol. Therefore, the correct answer choice will include enough waters to make the total mass of the water slightly higher than 106g. The molar mass of water is about 18 g/mol, so the correct answer includes 7 waters (7 x 18 = 126). This is A

7. MC Question 3 First, recognize this as a limiting reactant problem! 0.025L x 0.200 mol/L = 0.00500 mol BrO 3 - 0.030L x 0.500 mol/L = 0.0150 mol Br - According to the balanced equation, we need 5 times as much Br - as BrO 3 -. We have less than that, so Br - is limiting. Divide the moles of Br - by its own coefficient, then multiply by the coefficient of Br 2 : 0.0150 / 5 x 3 = 0.00900 mol Br 2 A

8. MC Question 4 Most of these answer choices contain the number 22.4, which is the volume in liters of one mole of gas at STP (standard molar volume). However, in this case we are given and asked for information about gases at the same temperature and pressure—there is no need to convert from liters to moles. Divide the liters of CO 2 by its own coefficient, then multiply by the coefficient of O 2 : 4.0 / 1 x 3 = 12 L O 2 B

9. MC Question 5 Since the empirical formula expresses the simplest possible whole number ratio of elements in a compound, divide each given number of moles by the smallest number of moles: K: 1.10/0.55 = 2 Te: 0.55/0.55 = 1 O: 1.65/0.55 = 3 The correct empirical formula is K 2 TeO 3. D

10. MC Question 6 The “before” picture shows 4 NH 3 molecules and 7 O 2 molecules. The balanced equation includes 4 NH 3 and 5O 2. Therefore, we are looking for an “after” picture that shows two excess O 2 molecules but otherwise matches what the balanced equation predicts—4 NO and 6 H 2 O. C Note that all of the other answer choices violate the law of conservation of mass!

11. MC Question 7 We can immediately eliminate answer choice A as ridiculous— the reaction clearly produces a solid, so there will be a precipitate. Beyond that, this is another limiting reactant problem. We begin with 0.050 mol MgCl 2 and 0.150L x 0.200mol/L = 0.030 mol NaOH. Since the ratio is 1:1, the NaOH is limiting. Therefore, all of the OH - will precipitate. Answer choice B is wrong. However, there will be some Mg 2+ left in solution from the excess MgCl 2. Answer choice D is wrong. C

12. MC Question 8 The precipitation reaction will be: Ag + + Br - → AgBr(s) Therefore, the number of moles of AgNO 3 required will be the same as the number of moles of Br - in the solution. 0.25 mol NaBr = 0.25 mol Br - 0.25 mol CaBr2 = 0.50 mol Br - 0.25 + 0.50 = 0.75 mol B

13. MC Question 9 The moles of nitric acid available to react must be found: 0.010L x 6.0 mol/L = 0.060 mol Silver is obviously in excess (10.0mol!), so the nitric acid is limiting. Divide moles of nitric acid by its own coefficient, then multiply by the coefficient of NO. 0.060 / 4 x 1 = 0.015 mol NO A

14. MC Question 10 First, find the number of moles of H 2 SO 4 in the described solution: 0.050L x 6.00 mol/L = 0.30 mol Now convert to grams using the given molar mass: 0.30 mol x 98.1 g/mol = a bit less than a third of 99 = a bit less than 33, so... C

15. 1. Free Response Q1

21. Free Response Q2