1. Conversion Factors Molar mass Molar mass atomic mass in g = 1 mole atomic mass in g = 1 mole Volume of gas at STP Volume of gas at STP 1 mole gas = 22.4L 1 mole gas = 22.4L Mole-mole ratio coefficients from balanced equation Mole-mole ratio coefficients from balanced equation Avogadro’s number Avogadro’s number 6.02 x 10 23 molecules = 1 mole 6.02 x 10 23 molecules = 1 mole

2. Stoichiometry g g mole X  mole Y mole X  mole Y L Molecules (gas at STP) Molar mass 22.4L=1 mol6.02 x 10 23 molecules = 1 mole For great ion review game www2.stetson.edu/mahjongchem/

3. 2AgNO 3 + H 2 S Ag 2 S + 2HNO 3 How many moles of Ag 2 S will be produced from 3.5 moles of AgNO 3 3.5 mole AgNO 3 x 1 mole Ag 2 S = 1.75 mole Ag 2 S 2 mole AgNO 3 2 mole AgNO 3 How many grams of HNO 3 will form when 2.5g of H 2 S react? 2.5g H 2 S x 1 mol H 2 S x 2 mol HNO 3 x 63.0g HNO 3 34.1g H 2 S 1 mol H 2 S 1 mol HNO 3 34.1g H 2 S 1 mol H 2 S 1 mol HNO 3 = 9.2g HNO 3

4. Molarity moles of solute L of solution Dilutions V 1 M 1 = V 2 M 2 Dilutions V 1 M 1 = V 2 M 2

5. How can we make 500 ml of 1.00 M HCl from 6.00 M HCl? V 1 = ? M 1 = 6.00 M V 2 =.500L M 2 = 1.00 M V 1 (6.00M) = (.500L)(1.00M) V 1 =.500L∙M =.0833L or 83.3ml 6.00M

6. What is the molarity of 250ml solution containing 9.46g CsBr? g  mole 9.46g x 1 mole CsBr = 0.0444 mole CsBr 213 g CsBr 213 g CsBr 250 ml =.250 L M = 0.0444 mole = 0.178 M CsBr.250L.250L

7. The amount of heat gained or lost depends on the amount of reactants used. 2Na 2 O 2 + 2H 2 O  4NaOH + O 2 + 215.76 kJ How much heat is released by the reaction of 5.0 moles of Na 2 O 2 ? 5.0 moles Na 2 O 2 x 215.76 kJ = 539.4 kJ 2 mole Na 2 O 2 2 mole Na 2 O 2 Heat of reaction (Q r ) is negative for exothermic reactions, but positive for endothermic reactions. So Q r = -539.4 kJ

8. The decomposition of potassium chlorate is an endothermic reaction. 2KClO 3 (s) + 89.5kJ  2KCl(s) + 3O 2 (g) How much heat is absorbed to decompose 1.50 grams of solid KClO 3 ? 1.50g KClO 3 x 1 mole KClO 3 x 89.5kJ 122.5 g KClO 3 2 mole KClO 3 122.5 g KClO 3 2 mole KClO 3 =.548kJ

9. Limiting Reactant 1 frame and 3 wheels are needed to create one tricycle If I have 10 frames and 12 wheels, how many tricycles can I make?

10. N 2 H 4 + 2H 2 O 2  N 2 + 4H 2 O Which is the limiting reactant when 0.750 moles of N 2 H 4 is mixed with 0.500 mole of H 2 O 2 ? If all the N 2 H 4 is used, how much H 2 O 2 would be needed? 0.750 mole N 2 H 4 x 2mole H 2 O 2 = 1.5 mole H 2 O 2 1 mole N 2 H 4 1 mole N 2 H 4 Do we have that much? No! So H 2 O 2 is limiting

11. N 2 H 4 + 2H 2 O 2  N 2 + 4H 2 O How much of the excess reactant (in moles) remains unchanged? 0.500 moles H 2 O 2 x 1 mole N 2 H 4 =.250 mole 2 mole H 2 O 2 2 mole H 2 O 2 0.750 mole -.250 mole =.500 mole N 2 H 4 How much H 2 O would be formed? 0.500 mole H 2 O 2 x 4 mole H 2 O = 1.00 mole H 2 O 2 mole H 2 O 2 2 mole H 2 O 2

12. % yield = actual x 100% % yield = actual x 100% theoretical theoretical If 5.50g of hydrogen reacts with nitrogen to form 20.4g of ammonia, what is the percent yield? N 2 (g) + 3H 2 (g)  2NH 3 (g) Actual = 20.4g Theoretical = 5.50g H 2 x 1 mol H 2 x 2mol NH 3 x 17.0g NH 3 2.0g H 2 3mol H 2 1mol NH 3 2.0g H 2 3mol H 2 1mol NH 3

13. % yield = actual x 100% % yield = actual x 100% theoretical theoretical % yield = 20.4g x 100% = 65.4% % yield = 20.4g x 100% = 65.4% 31.2g 31.2g

14. What is the molecular formula of the molecule that has an empirical formula of CH 2 O and a molar mass of 120.12 g/mol? What is the molecular formula of the molecule that has an empirical formula of CH 2 O and a molar mass of 120.12 g/mol? Empirical formula mass: 12.0 + 2.0 + 16.0 = 30.0 amu X = 120.12 amu = 4 30.0 amu 30.0 amu So C 4 H 8 O 4 is the molecular formula.

15. Determining empirical formula Find the empirical formula of a compound that contains 53.7% iron and 46.3% sulfur. Find the empirical formula of a compound that contains 53.7% iron and 46.3% sulfur. % composition  mass of 100 g sample  moles  mole ratio % composition  mass of 100 g sample  moles  mole ratio 53.7g x 1 mol/55.8g =.962 mol Fe 53.7g x 1 mol/55.8g =.962 mol Fe 46.3g x 1 mol/32.1g = 1.44 mol S 46.3g x 1 mol/32.1g = 1.44 mol S.962/.962 = 1, 1.44/.962 = 1.5.962/.962 = 1, 1.44/.962 = 1.5 1:1.5 ratio x 2 = 2:3 ratio 1:1.5 ratio x 2 = 2:3 ratio Fe 2 S 3 Fe 2 S 3