Stoichiometry Chapter 11 (page 326)

Essential Question!! Why do scientist compare mole ratios, and why is it important in chemical reactions??

Vocabulary: Section 1 Stoichiometry Moles Stoichiometric equivalent Mole ratio Gram to gram calculations

2CO(g) + O2(g) 2CO2(g) Chemical equations tell stories… But what exactly do they tell us? 2CO(g) + O2(g) 2CO2(g) They tell us what compounds we start with: Carbon monoxide (CO) gas Oxygen (O2) gas what compounds are formed: Carbon dioxide (CO2) gas

They tell us how much of each compound is involved Chemical equations tell stories… What else do they tell us? 2CO(g) + O2(g) 2CO2(g) 2 CO molecules 1 O2 molecules 2 CO2 molecules They tell us how much of each compound is involved stoichiometry: the study of the amounts of substances involved in a chemical reaction.

2CO(g) + O2(g) 2CO2(g) 2 CO molecules 2 dozen CO molecules 2 moles CO molecules 2 x (6.023 x 1023) CO molecules 2 CO2 molecules 2 dozen CO2 molecules 2 moles CO2 molecules 2 x (6.023 x 1023) CO2 molecules 1 O2 molecules 1 dozen O2 molecules 1 mole O2 molecules (1 x) 6.023 x 1023 O2 molecules

Number of moles is not conserved Is that okay? 2CO(g) + O2(g) 2CO2(g) Number of moles is not conserved 2 moles CO molecules + 1 mole O2 molecules ≠ 2 moles CO2 molecules

2CO(g) + O2(g) 2CO2(g) = + = This chemical equation is balanced 2 C atoms 2 O atoms 2 O atoms 2 C atoms 4 O atoms Number of atoms is conserved = + = 2CO(g) + O2(g) 2CO2(g) This chemical equation is balanced

Coefficients are important 1 bag cake mix + 3 eggs + ¼ cup oil + 1 cup water 1 batch cupcakes Write as a ratio:

Coefficients are important Fermentation of sugar (glucose) into alcohol: C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g) 1 mole glucose 2 moles ethanol 2 moles carbon dioxide 1 moles 3 moles 7.5 moles 2 moles 6 moles 15 moles x 3 ethanol/ carbon dioxide x 7.5 glucose will yield

Fermentation of sugar (glucose) into alcohol: C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g) 1 mole glucose 2 moles ethanol 2 moles carbon dioxide Write as a ratio: mole ratio: a ratio comparison between substances in a balanced equation. It is obtained from the coefficients in the balanced equation.

Mole ratios Consider the following equation: CO(g) + 2H2(g) CH3OH(l) carbon monoxide hydrogen methanol Compare the reactant CO to the product CH3OH.

Stoichiometry 1. Mass of substance A 3. Amount in mol of substance B Convert by dividing by the molar mass of A B 2. Amount in mol of Substance A Convert using the mol ratio A Convert by mult. by molar mass of B 4. Mass of Substance B given in the chemical equation

Process for calculating grams from grams given

Method Mass - Mass mass A mass B = coeff A X FM A coeff B X FM B

Method Mass - Volume mass A vol B = coeff A * FM A coeff B * 22.4

Method Volume - Volume vol B Vol A * coeff B coeff A

Assignment Take a new sheet of paper and fold it into three sections Write your name, the title of the chapter and the number On the first section from the sheet of paper, please write six things that you learned from your notes so far that could appear on your test.

1. Given the equation N2 + 3 H2  2NH3 If 4 1.Given the equation N2 + 3 H2  2NH3 If 4.00 mol of H2 react, how many mol of NH3 will be produced? 2 2.66 mol 4.0 mol X = 3

How many moles of sodium will react with water to produce 4 How many moles of sodium will react with water to produce 4.00 mol of Hydrogen? 2 Na + 2H2O  2 NaOH + H2 2 8 mol sodium 4 mol H2 X = 1

How many moles of H2SO4 will react with 18. 0 mol of Al How many moles of H2SO4 will react with 18.0 mol of Al? 2 Al + 3 H2SO4  Al2(SO4)3 + 3 H2 3 18.0 mol Al X = 27.0 moles 2

What mass of KClO3 do you need to produce 0. 50 mol of O2 What mass of KClO3 do you need to produce 0.50 mol of O2? 2 KClO3  2 KCl + 3 O2 2 0.50 mol X 0.33 mol KClO3 = 3 0.33 mol KClO3 X 122.6 g/mol = 40.866 g

What mass of Zn metal do you need to produce 0. 50 mol of ZnCl2 What mass of Zn metal do you need to produce 0.50 mol of ZnCl2? Zn (s) + 2HCl  ZnCl2(s) + H2(g) 1 0.50 mol X 0.5 Zn = 1 0.5 mol Zn X 65.4 g/mol = 32.7 g Zn

What mass of NH3 do you need to produce 0. 25 mol of H2 What mass of NH3 do you need to produce 0.25 mol of H2? N2 + 3H2(g)  2 NH3 3 0.25 mol X 0.375 H2 = 2 0.375 mol H2 X 0.75 g/mol = 0.75 g H2

Assignment Write a three dollar summary of this section (based on what you learned) and using three vocabulary words learned Answer questions # 1- 3 on page 360 Turn in completed work Honors Chemistry Homework: Page 362 # 38 - 45

Vocabulary: Section 2 Percent yield Actual yield Theoretical yield

+ In theory, all 100 kernels should have popped. Did you do something wrong? + 100 kernels 82 popped 18 unpopped

Percent yield What you get to eat! + 100 kernels 82 popped 18 unpopped

actual yield: the amount obtained in the lab in an actual experiment. theoretical yield: the expected amount produced if everything reacted completely.

Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?) - There is usually some human error, like not measuring exact amounts carefully - Maybe the heating time was not long enough; not all the Na2HCO3 reacted - Maybe Na2CO3 was not completely dry; some H2O(l) was measured too - CO2 is a gas and does not get measured

Percent yield in the lab Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Let’s calculate the percent yield obtained in experiment calculated

Conversions of Mass to Amounts In Moles Example

Mass to Mass Calculations Example

Assignment Write a three dollar summary of this section (based on what you learned) Answer questions # 4- 6 on page 360 Turn in completed work Honors Chemistry Homework Page 363 # 46 - 57

Vocabulary: 3 Limiting reactant Excess reactant

Recipe: Suppose you want to make 2 ham & cheese sandwiches 4 slices of bread 4 slices of ham 2 slices of cheese

No, you are limited by the cheese! Suppose you want to make 2 ham & cheese sandwiches Can you still make 2 ham & cheese sandwiches if you have 4 slices of bread, 4 slices of ham, and 1 slice of cheese? Limiting factor No, you are limited by the cheese! You can only get 1 ham & cheese sandwich.

Limiting Reactants A limiting reactant is the reactant that limits the amount of the other reactants that can combine to form product in a chemical reaction An excess reactant is the reactant that is left over after the other reactant runs out, or the reaction is completed

For a chemical reaction: Reactant A is in excess so the reaction will stop when you run out of reactant B. Reactant B is the limiting reactant. The amount of product C will depend on how much reactant B is present. Excess reactant

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s) 1. What is the limiting reactant? 150 g 60 g Step 1: Convert masses to moles Step 2: Use mole ratios to find the limiting reactant

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s) 1. What is the limiting reactant? 150 g 60 g Step 1: Convert masses to moles available Fe2O3 available Al Step 2: Use mole ratios to find the limiting reactant needed to react with all Fe2O3 needed to react with all Al

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s) 1. What is the limiting reactant? 150 g 60 g Step 1: Convert masses to moles available Fe2O3 available Al Step 2: Use mole ratios to find the limiting reactant needed to react with all Fe2O3 needed to react with all Al

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s) 1. What is the limiting reactant? 150 g 60 g available Fe2O3 There is not enough Fe2O3 available to react with all the Al, so Fe2O3 is the limiting reactant needed to react with all Al

Example # 1 & 2

The Haber-Bosch process for the synthesis of ammonia: N2(g) + 3H2(g) → 2NH3(s)

The Haber-Bosch process for the synthesis of ammonia: N2(g) + 3H2(g) → 2NH3(s)

Vocabulary: Section 4 Excess reactant Percent yield

What is left? 6Li(s) + N2(g) → 2Li3N(s) Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of excess reactant left Given: N2 is the excess reactant 1.15 moles N2 (need) 1.66 moles N2 (have) Relationships: 28.01 g/mole N2 Solve: 1) How many moles N2 remain? 2) Convert moles to grams Answer: 14 g of N2 will remain at the end of the reaction.

Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: 95.61 g/mole CuS 1 mole CuSO4 ~ 1 mole CuS Solve: 1) How many moles of CuS? 2) How many moles of CuSO4? 3) What is the concentration of CuSO4? Have:

Assignment On the second section of that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.

Percent Yield Theoretical yield is the maximum amount of product that can be produced from a given amount of reactant Actual yield is the measured amount of product obtained from a reaction Percent yield is the ratio of the actual yield to the theoretical yield

Assignment Write a three dollar summary of this section (based on what you learned) and using the vocabulary learned this section Answer questions # 7- 8 on page 360 Turn in completed work Honors Chemistry Homework Page 364 # 58 - 63

Test:- Next Tuesday or Thursday depending on your class Homework requirement: page 360 # 1 - 8 Make sure you have all work assigned between pages 360 and 363 completed and turned in by your test date.