2. Stoichiometry Part 3!

3. Mass to Mole Conversions Remember you CANNOT convert directly from mass! If you begin with mass, you must convert to moles FIRST! Then compare moles to moles using coefficients. Mass → conversion to moles → mole ratio

4. Sample Problem 9-4 The catalytic oxidation of ammonia is run using 824 g of ammonia and excess oxygen. How many moles of NO are formed? First, write the balanced equation:  NH 3 + O 2 → NO + H 2 O  4NH 3 + 5O 2 → 4NO + 6H 2 O

5. 4NH 3 + 5O 2 → 4NO + 6H 2 O Convert from 824 g NH 3 to moles:  824 g NH3 x 1 mole NH 3 = 48.3839525 mole 17.03044 g Use the mole ratio to find moles of NO:  48.3839525 mole NH 3 x 4 mole NO = 48.3839525 4 mole NH 3  Check significant digits! 48.4 mole NO

6. 4NH 3 + 5O 2 → 4NO + 6H 2 O How many moles of H 2 O are formed from 824 g of NH 3 ? 824 g NH 3 = 48.4 mole NH 3 Then use mole ratio:  48.4 mole NH3 x 6 mole H 2 O = 72.57592875 mole 4 mole NH 3  Correct for significant digits: 72.6 mole H 2 O

7. Mass-mass calculations First convert from mass to moles Then use mole ratio Then convert from moles to mass Remember, you CANNOT convert directly from mass of one substance to mass of another!

8. Sample problem 9-5 The reaction of tin with hydrogen fluoride produces tin (II) fluoride and hydrogen gas How many grams of SnF 2 are produced from the reaction of 30.00 g of HF with Sn? First, write the balanced equation:  Sn + HF → SnF 2 + H 2  Sn + 2HF → SnF 2 + H 2

9. Sn + 2HF → SnF 2 + H 2 30.00 g HF x 1 mole HF = 1.499557 mole HF 20.006 g HF 1.499557 mole HF x 1 mole SnF 2 = 0.74978 2 mole HF mole SnF 2 0.74978 mole SnF 2 x 156.706 g SnF 2 =117.49 1 mole SnF 2 g SnF 2 Correct for significant digits: 117.5 g SnF 2

10. Now it's your turn! Do practice problems 1-2 page 285 and practice problems 1-3 page 287.