1. Chapter 12: Chemical Quantities Section 12.2: Using Moles

2. Used to relate moles of one substance to moles of another substance Use coefficients in the equation to convert to moles of other reactants or products Balanced chemical equations

3. The following recipe serves 4: 4 potatoes (.3 lb each) 2 onions (.2 lb each) 8 carrots (.1 lb each) 4 stalks of celery (.05 lb each) 1.4 lbs of water What is the total mass of the soup? How would you make enough for 8? How about 1240? **With a balanced chemical equation and number of moles, we can predict the exact amount of reactant and product in a reaction**

4. There are 4 steps to follow: 1) Write the balanced chemical equation 2) Convert the given mass or volume to moles 3) Use the coefficients in the chemical equation to set up a mole ratio (the coefficients are the # of moles!) HINT: The substance you are solving for goes on TOP 4) Convert these moles back to mass or volume as required

5. Use: Grams A ↔ Moles A ↔ Moles B ↔ Grams B molar mass coefficients molar mass Steps

6. N 2 + 3H 2  2NH 3 This reaction can be stated as: 1 mole of nitrogen will react with 3 moles of hydrogen to produce 2 moles of ammonia.

7. #10) What mass of CO 2 forms when 95.6 g of C 3 H 8 burns? C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) 3 C = 36.033 gC= 12.011g 8 H= 8.064 g 2 O = 32 g C 3 H 8 = 44.097 g/mol CO 2 = 44.011 g/mol 95.6 g C 3 H 8 x 1 mol C 3 H 8 x 3 mol CO 2 x 44.011 g CO 2 = 44.097g C 3 H 8 1 mol C 3 H 8 1 mol CO 2 = 286.2 g CO 2 Practice Problems (pg. 415)

8. #11) How many grams of fluorine are required to produce 10 g XeF 6 ? Xe (g) + 3F 2 (g) → Xe F 6 (s) F 2 : 2 x 18.998 g = 37.996 g Xe F 6 : 131.29 + (6 x 18.998) = 245.278 g 10 g XeF 6 x 1 mol XeF 6 x 3 mol F 2 x 37.996 g F 2 = 245.278 g XeF 6 1 mol XeF 6 1 mol F 2 = 4.65 g F 2 Practice (cont)

10. The reactant that runs out first in a reaction/ stops the reaction LIMITING REACTANT

11. #1) What is the limiting reactant in producing water (H 2 O), 5 g H 2 or 5 g of O 2 ? (convert g reactant → mol of product) 2H 2 + O 2 → 2H 2 O 5 g H 2 x 1 mol H 2 x 2 mol H 2 O = 2.48 mol H 2 O 2.0158 g H 2 2 mol H 2 5 g O 2 x 1 mol O2 x 2 mol H2O = 0.31 mol H 2 O 31.998 g O2 1mol O 2 Limiting reactant is O 2 (runs out first) Practice Problems

12. 2) Which is the limiting reactant in producing NH 3, 3.75 g N 2 or 3.75 g H 2 ? N 2 + 3H 2 → 2 NH 3 3.75g H 2 x 1 mol H 2 x 2 mol NH 3 = 1.24 mol NH 3 2.0158 g H 2 3 mol H 2 3.75g N 2 x 1 mol N 2 x 2 mol NH 3 = 0.27 mol NH 3 28 g N 2 1 mol N 2 Limiting reactant is N 2 Practice (cont)