1. Stoichiometry 2: grams to grams Chemistry 2012-2013 Ms. Boon 10.18 & 10.19

2. Catalyst

3. Helpful materials for today

4. Mass (grams) of COMPOUND A Mass (grams) of COMPOUND B Amount (Moles) of COMPOUND A # particles (atoms or molecules) of COMPOUND A Amount (Moles) of COMPOUND B # particles (atoms or molecules) of COMPOUND A Divide by molar mass (g/mol) of compound A Divide by Avogadro’s # (6.02 x 10 23 ) Multiply by Avogadro’s # (6.02 x 10 23 ) Multiply by molar mass (g/mol) of compound B Multiply by mole ratio from balanced equation Mol Compound B Mol Compound A

5. What are we converting? Look for clues in the question! Na 2 O + H2O  2NaOH 1. How many moles of Na 2 O are needed to produce 22 moles of NaOH? A: moles of NaOH to moles of Na 2 O 2. How many grams of Na 2 O are needed to produce 30 moles of NaOH? A: moles of NaOH to grams of Na 2 O 3. How many moles of Na 2 O are needed to produce 200g of NaOH? A: grams of NaOH to moles of Na 2 O 4. How many grams of Na 2 O are needed to produce 58 g of NaOH? A: grams of NaOH to grams of Na 2 O

6. Quick Check WW hat type of conversion is this question? HH ow many grams of oxygen gas (O 2 ) can be made from the decomposition of 136g of hydrogen peroxide (H 2 O 2 )? 22 H 2 O 2  22 H 2 O + O 2 VV ote for your answer choice! 1. Grams of O 2 to grams of H 2 O 2 ? 2. Moles of O 2 to grams of H 2 O 2 ? 3. Grams of H 2 O 2 to grams of O 2 ? 4. Grams of H 2 O 2 to moles of O 2 ?

7. Activity: Stoichiometry Puzzle  You and your partner will receive a 12 puzzle pieces with stoichiometry problems on them.  Your task is to sort the problems into four piles according to the type of conversion in the problem.  The four types of conversions are:  Gram to gram  Mole to mole  Gram to mole  Mole to gram  When you think you have sorted the problems correctly, raise your hand to get your answers checked.

8. Amount (Moles) of COMPOUND A Amount (Moles) of COMPOUND B Multiply by mole ratio from balanced equation Mol Compound B Mol Compound A We worked on this part earlier this week. Mole to Mole Conversions are part of every stoichiometry problem.

9. Mass (grams) of COMPOUND A Mass (grams) of COMPOUND B Amount (Moles) of COMPOUND A Amount (Moles) of COMPOUND B Divide by molar mass (g/mol) of compound A Multiply by molar mass (g/mol) of compound B Multiply by mole ratio from balanced equation Mol Compound B Mol Compound A Today we are adding gram to gram conversions. We already know how to do all the steps. Today we are putting them together.

10. Mass (grams) of COMPOUND A Mass (grams) of COMPOUND B Amount (Moles) of COMPOUND A Amount (Moles) of COMPOUND B Divide by molar mass (g/mol) of compound A Multiply by molar mass (g/mol) of compound B Multiply by mole ratio from balanced equation Mol Compound B Mol Compound A Example 1: How many grams of oxygen gas (O 2 ) can be made from the decomposition of 136g of hydrogen peroxide (H 2 O 2 )? 2 H 2 O 2  2 H 2 O + O 2 136g H 2 O 2 Divide by 34 g/mol 4 mol H 2 O 2 1 mol O 2 2 mol H 2 O 2 2 mol O 2 Multiply by 32 g/mol 64g O 2

11. Mass (grams) of COMPOUND A Mass (grams) of COMPOUND B Amount (Moles) of COMPOUND A Amount (Moles) of COMPOUND B Divide by molar mass (g/mol) of compound A Multiply by molar mass (g/mol) of compound B Multiply by mole ratio from balanced equation Mol Compound B Mol Compound A Example 2: How many grams of hydrogen peroxide (H 2 O 2 ) are needed to produce 180g water (H 2 O)? 2 H 2 O 2  2 H 2 O + O 2 180g H 2 O Divide by 18 g/mol 10 mol H 2 O 2 mol H 2 O 2 2 mol H 2 O 10 mol H 2 O 2 Multiply by 34 g/mol 340g H 2 O 2

12. Amount (Moles) of COMPOUND A Amount (Moles) of COMPOUND B Multiply by mole ratio from balanced equation Mol Compound B Mol Compound A Mole to Mole Conversion Example: How many moles aluminum are needed if 3 mol Fe 2 O 3 completely react.Fe 2 O 3 + 2 Al→ 2 Fe + Al 2 O 3 Converting from mol Fe 2 O 3 to mol Al 2 mol Al____ 1 mol Fe 2 O 3 X 3 mol Fe 2 O 3 = 6 mol Al

13. Mass (grams) of COMPOUND A Mass (grams) of COMPOUND B Amount (Moles) of COMPOUND A Amount (Moles) of COMPOUND B Divide by molar mass (g/mol) of compound A Multiply by molar mass (g/mol) of compound B Multiply by mole ratio from balanced equation Mol Compound B Mol Compound A Gram to Gram Conversions Example: How many grams of BrCl form when 140g Cl 2 react with excess Br 2 ? Br 2 + Cl 2 → 2BrCl 2 mol BrCl 1 mol Cl 2 X 2 mol Cl 2 = 4 mol BrCl 140g Cl 2 ÷ 70 g/mol = = X 115 g/mol 460 g BrCl

14. Mass (grams) of COMPOUND A Amount (Moles) of COMPOUND A Amount (Moles) of COMPOUND B Divide by molar mass (g/mol) of compound A Multiply by mole ratio from balanced equation Mol Compound B Mol Compound A Gram to Mole Conversion Example: How many moles of BrCl form when 210g Cl 2 react with excess Br 2 ? Br 2 + Cl 2 → 2BrCl 2 mol BrCl 1 mol Cl 2 X 3 mol Cl 2 6 mol BrCl 210g Cl 2 ÷ 70 g/mol = =

15. Mass (grams) of COMPOUND B Amount (Moles) of COMPOUND A Amount (Moles) of COMPOUND B Multiply by molar mass (g/mol) of compound B Multiply by mole ratio from balanced equation Mol Compound B Mol Compound A Mole to Gram Conversion Example: How many grams of BrCl form when 5 mol Cl 2 react with excess Br 2 ? Br 2 + Cl 2 → 2BrCl 2 mol BrCl 1 mol Cl 2 X 5 mol Cl 2 = 10 mol BrCl = X 115 g/mol 1150 g BrCl

16. Exit Slip (3e) 10.18 & 19 N 2 + 3 H 2  2 NH 3 1. What is the mole ratio of H 2 to NH 3 ? 2. How many moles of H 2 are needed to produce 10 moles NH 3 ? 3. How many moles H 2 are needed to produce 170 grams NH 3 ? (molar mass of NH 3 is 17g/mol) 4. How many grams H 2 are needed to produce 170 grams NH 3 ? (NH 3 =17g/mol; H 2 = 2g/mol) Challenge: If you have 1 mol N 2 and 6 mol H 2, how many moles of NH 3 are produced? Why? Tonight’s HW: Read pp. 306-307, practice problems #1-4 on p. 307; problems p. 334 #1 & 3