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Stoichiometry L. What is it really? Cooking Method of finding – how many cookies you can make given these ingredients – how many ingredients you had.

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Presentation on theme: "Stoichiometry L. What is it really? Cooking Method of finding – how many cookies you can make given these ingredients – how many ingredients you had."— Presentation transcript:

1 Stoichiometry L

2

3 What is it really? Cooking Method of finding – how many cookies you can make given these ingredients – how many ingredients you had to make this many cookies

4 All goes back to balanced equations First write the equation and balance it. You are observing the combustion of methane, CH 4. Your teacher asks you how many moles of water are expected to form in the complete combustion of 2.50 mol of methane. CH 4 + O 2  CO 2 + H 2 OUNBALENCED CH 4 + 2O 2  CO 2 + 2H 2 OBALENCED

5 Now the math Remember that the balanced equations shows the relationship of how many moles of each compound is used in this reaction – Like the ingredients on a recipe (3 eggs + ½ cup water = eggy water) So the coefficient of one compound to the next is the mole ratio

6 Now the math (….continued) You are observing the combustion of methane, CH 4. Your teacher asks you how many moles of water are expected to form in the complete combustion of 2.50 mol of methane. CH 4 + 2O 2  CO 2 + 2H 2 O So if we had 1 mole of methane we would produce 2 moles of water But since we have 2.50 mol CH 4 …

7 The math You are observing the combustion of methane, CH 4. Your teacher asks you how many moles of water are expected to form in the complete combustion of 2.50 mol of methane. CH 4 + 2O 2  CO 2 + 2H 2 O Therefore we would expect to retrieve 5 mols of H 2 O from the combustion of 2.5 mol of methane 5 mol H 2 O Mole ratio

8 mole to mole The synthesis reaction between nitrogen monoxide and oxygen forms nitrogen dioxide. – How many moles of nitrogen monoxide would be needed to fully react with 3.6 moles of oxygen – How many moles of nitrogen dioxide would be produced from 3.6 moles of oxygen – How many moles of oxygen would be needed to produce 4.67 mol nitrogen dioxide? – How many moles of nitrogen monoxide would be needed to produce 4.67 mol nitrogen dioxide?

9 Building off that… If you can do mol to mol, the rest should be child’s play Mass to mol – Convert mass to mol, then mol to mol. – The end Mass to mass – Convert mass of given to mol, convert mols of given to moles of wanted, convert mol to mass – The end Particles to anything – Use Avagotto’s number (6.022 x particles / mol )

10 Mass to mol In a reaction between the elements aluminum and chlorine, aluminum chloride is produced. How many grams of AlCl3 will be produced if 2.50 moles of Al react? Al + Cl 2  AlCl 3 Unbalanced 2Al + 3Cl 2  2AlCl 3 Balanced

11 Mol to mass continued 2Al + 3Cl 2  2AlCl 3 How many grams of AlCl 3 will be produced if 2.50 moles of Al react? Therefore g of AlCl 3 will be produced by the reaction of 2.50 mol of Al 2.50 mol AlCl 3 1 mol AlCl g AlCl 3

12 Practice The ammonia (NH 3 ) used to make fertilizers for lawns and gardens is made by reacting nitrogen and hydrogen. – Determine the mass in grams of NH 3 formed from 1.34 moles of nitrogen. – What is the mass in grams of hydrogen required to react with 1.34 moles of nitrogen? – How many moles of nitrogen are required to produce 11.7 grams of NH 3 ?

13 Mass to mass When nitrogen and hydrogen react, they form ammonia gas, which has the formula NH 3. If 56.0 g of nitrogen are used up in the reaction, how many grams of ammonia will be produced? N 2 + H 2  NH 3 Unbalanced N 2 + 3H 2  2NH 3 Balanced

14 When nitrogen and hydrogen react, they form ammonia gas, which has the formula NH 3. If 56.0 g of nitrogen are used up in the reaction, how many grams of ammonia will be produced? N 2 + 3H 2  2NH 3 Convert grams to mol Convert mol to gram Therefore, g of NH3 are produced from 56 g of N mol NH g NH 3

15 Particles to Particles Hydrogen and Oxygen react to produce water. How many particles of hydrogen would be needed to react fully with 1.01 x particles of oxygen? H 2 + O 2  H 2 O Unbalanced 2H 2 + O 2  2H 2 OBalanced

16 Hydrogen and Oxygen react to produce water. How many particles of hydrogen would be needed to react fully with 1.01 x particles of oxygen? 2H 2 + O 2  2H 2 O Convert particles to mol Convert mol to mol Convert mol to particles Therefore we expect to need 2.14 x particles of H 2 to react fully with 1.01 x particles of O x mol O x particles of H 2

17 Practice Packet!

18 Percent yield How much are you actually getting out of your experiment? % Yield = Actual yield / theoretical yield x 100

19 What is the total mass of H 2 O produced when 384g Cu is completely consumed in the following reaction? Cu + 4HNO 3  Cu(NO 3 ) 2 + 2H 2 O + 2NO 2 Convert mass to mol Convert mol to mol Convert mol to mass Therefore we expect to produce 216g of H 2 O. What is the percent yield if this procedure was carried out in the lab and only produced 200g of H 2 O 216g H 2 O

20 We expected to produce 216g of H 2 O. What is the percent yield if this procedure was carried out in the lab and only produced 200g of H 2 O What was the actual yield? – 200g What was the theoretical yield? – 216g Solve Therefore, our % yield is 92.6% % yield = Actual yield Theoretical yield 200g 216g 92.6% =

21 Practice Ethanol (C 2 H 5 OH) is produced from the fermentation of sucrose in the presence of enzymes: C 12 H 22 O 11(aq) + H 2 O (g)  C 2 H 4 OH (l) + CO 2(g) determine the theoretical and percent yields of ethanol if 684g sucrose undergoes fermentation and 349g ethanol is obtained T.Y. = 369g % yield = 94.6% Lead (II) oxide is obtained by roasting galena, lead (II) sulfide, in air: PbS (s) + O2 (g)  PbO (s) + SO 2(g) Determine the theoretical and percent yields of PbO if 200.0g PbS is heated and 170.0g PbO is obtained T.Y. = 186.6g PbO % Yield = 91.10%

22 Limiting reagent The reactant that is completely consumed during a rxn Reactants that remain are excess reactants

23 To find the limiting reagent Compare the actual mol:mol ratio to the balanced formula’s mol:mol ratio Calculations must be based on given amounts of limiting reactants

24 Fe + 2NiO(OH) + 2H 2 O  Fe(OH) 2 + 2Ni(OH) 2 Determine the number of moles of Fe(OH) 2 produced if 5.00 mol Fe and 8.00 mol of NiO(OH) react. What is the limiting reactant? What reactant is in excess and by how many moles? 1mol Fe::2mol NiO(OH). Therefore,5mol Fe would need 10mol NiO(OH) Since we only have 8.00mol NiO(OH), this is our limiting reagent.

25 Fe + 2NiO(OH) + 2H 2 O  Fe(OH) 2 + 2Ni(OH) 2 Determine the number of moles of Fe(OH) 2 produced if 5.00 mol Fe and 8.00 mol of NiO(OH) react. What is the limiting reactant? What reactant is in excess and by how many moles? Since NiO(OH) is the limiting reagent, find how many moles of Fe are needed to completely react with the amount of it we do have 4 Mols of Fe are needed Therefore 1 mol of Fe is in excess 4mol Fe 5.00mol Fe (provided) - 4mol Fe (needed) = 1mol Fe (excess)

26 It gets complicated If you are given anything other then mol, you need to first convert to mol for BOTH reagents before you can determine the limiting factor!

27 Zn + 2MnO 2 +H 2 O  Zn(OH) 2 + Mn 2 O 3 Determine the limiting reactant if 25.0g Zn and 30.0g MnO 2 are used, Determine the mass of Zn(OH) 2 produced. 0.77g moles of MnO 2 required to react with 25.0g of Zn 0.17 moles of Zn required to react with 30.0g of Mn(OH) 2 Mn(OH) 2 is the limiting reagent – excess of 0.21mol Zn ( = 0.21) Use 30.0g to find the mass of Zn(OH) g Zn(OH) 2


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