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Chapter 9 STOICHIOMETRY. What’s it mean? Greek stoikheion, meaning elementelement metron, meaning measuremeasure In English….chemical recipe.

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Presentation on theme: "Chapter 9 STOICHIOMETRY. What’s it mean? Greek stoikheion, meaning elementelement metron, meaning measuremeasure In English….chemical recipe."— Presentation transcript:

1 Chapter 9 STOICHIOMETRY

2 What’s it mean? Greek stoikheion, meaning elementelement metron, meaning measuremeasure In English….chemical recipe

3 IDEAL STOICHIOMETRY Assumes a reaction goes to completion and that all reactants will turn into products.

4 STEPS to SOLVING PROBLEMS 1. Write and balance the chemical reaction. The coefficients will give the mole ratio (the recipe). 2. Find the question or unknown. ?____ = given x conversion x mole mole 3. Start with the given and use conversions to reach the unknown. Every problem will include a mole ratio step.

5 Go with the flow…. chart to solve the 4 simple types of problems ^ (use coefficients from balanced reaction) Grams of given Molar mass Moles given Mole ratio Moles unknown Molar mass Grams unknown

6 1. MOLE – MOLE What are we calculating? Moles of any product or reactant, given moles of one other product or reactant.

7 EX. How many moles of ammonia are produced when 6 moles of hydrogen gas react with an excess of nitrogen gas? Balanced reaction: 3H 2 (g) + N 2(g)  2NH 3 (g) ^moles ^mole ^moles ? Moles NH 3 = 6 moles H 2 x 2 moles NH 3 = 4 moles NH moles H 2

8 Ex. How many moles of nitrogen are needed for the reaction to go to completion? ? Moles N 2 = 6 moles H 2 x 1 mole N 2 = 2 moles N moles H 2

9 2. MOLE - MASS What are we calculating? Grams of any product or reactant, given moles of any product or reactant.

10 Ex. When magnesium burns in air, it combines with oxygen to form magnesium oxide. What mass of magnesium oxide is produced from 2.00 moles of magnesium? Balanced Reaction: 2 Mg (s) + O 2 (g)  2MgO (s) ?g MgO = 2 moles Mg x 2 moles MgO x g MgO = 80.6 g MgO 1 2 moles Mg 1 mole MgO

11 3. MASS – MOLE What are we calculating? Moles of any product or reactant, given grams of any product or reactant. Ex. How many moles of mercury (II) oxide are needed to produce 125 grams of oxygen in a decomposition reaction? Balanced Reaction: 2 HgO (s)  O 2 g) + 2Hg (l) ?moles HgO= 125g O 2 x 1mole O 2 x 2 moles HgO = 7.81 moles HgO g O 2 1 mole O 2

12 4. MASS - MASS What are we calculating? Grams of any product or reactant, given grams of any product or reactant. Ex. Nitrous oxide is sometimes used as an anesthetic in dental work. It is produced when ammonium nitrate is decomposed into dinitrogen monoxide and water. How many grams of ammonium nitrate are needed to produce 33.0 g of N 2 O? Balanced Reaction: NH 4 NO 3(s)  N 2 O (g) + 2H 2 O (l) ?g NH 4 NO 3 = 33.0g N 2 O x 1 mole N 2 O x 1 mole NH 4 NO 3 x 80.06g NH 4 NO 3 = 60.0g NH 4 NO g N 2 O 1 mole N 2 O 1 mole NH 4 NO 3

13 How many grams of water are produced during the reaction? (Law of Conservation = 60.0 g – 33.0 g) ?g H 2 O = 33.0 g N 2 O x 1 mole N 2 O x 2 moles H 2 O x g H 2 O = 27.0 g H 2 O g N 2 O 1 mole N 2 O 1 mole H 2 O

14 MORE EXAMPLES…. Acetylene gas (C2H2) is used in welding and produces a very hot flame when burned in pure oxygen. How many grams of each product are produced when 25 kg of acetylene burns completely? Balanced Reaction: C 2 H 2 + 5/2 O 2  2 CO 2 + H 2 O 2C 2 H O 2  4 CO H 2 O ? gCO 2 = 25kg x 10 3 x 1 moleCO 2 x 4 moleCO 2 x 44.01gCO 2 = g CO 2 1 1kg 26.04g C 2 H 2 2 moleC 2 H 2 1 mole CO 2 ?g H 2 O= 25kg x 10 3 x 1 moleCO 2 x 2 mol H 2 O x g H 2 O = 1.7 x 10^4g 1 1kg 26.04g C 2 H 2 2 moleC 2 H 2 1 mole H 2 O H 2 O

15 MOLAR VOLUME of a GAS Rather than weighing gaseous products and reactants, it is often easier to consider the VOLUME of gas reacted or produced. For this to be useful in a stoichiometry problem, we must relate the volume of a gas to a number of moles. MOLAR VOLUME OF A GAS Every gas has volume = 22.4 L at STP STP = standard temperature and pressure 1 atm and 0.0 degrees C

16 Ex. Using molar volume Acetylene burns in oxygen. If L of acetylene is burned at STP, what volume of carbon dioxide is produced? 2 C 2 H O 2  4 CO 2 + 2H 2 O ?L CO 2 = 100.0L C 2 H 2 x1 mol C 2 H 2 x 4 mol CO 2 x22.4 L CO 2 = LC 2 H 2 2 mol C 2 H 2 1 mol CO 2 L CO 2 How many moles of water will be produced? ?mol H 2 O = 100.0L C 2 H 2 x1 mol C 2 H 2 x 2 mol H 2 O = mol H 2 O LC 2 H 2 2 mol C 2 H 2

17 If 84 L of acetylene are used, how many liters of oxygen are needed for complete combustion? ? L O 2 = 84 L C 2 H 2 x 1 mol C 2 H 2 x 5 mol O 2 x 22.4L O 2 = 210 L O L C 2 H 2 2 molC 2 H 2 1 mol O 2

18 PERCENT YIELD and STOICHIOMETRY THEORETICAL YIELD The maximum amount of product that can be produced during a reaction. Predicted by stoichiometry and ideal conditions. ACTUAL YIELD How much product is really produced during the reaction. A percent yield calculation tells us how efficient a reaction is.

19 % YIELD % yield = actual yield x 100 theoretical % yield + % error = 100%

20 % yield examples A reaction is supposed to produce 200. grams of HCl. If it actually yield 178 grams of HCl, how efficient is the reaction? % yield = actual x 100 = 178g x 100 = 89.0 %yield ; 11 % error theoretical 200g

21 If chlorine and hydrogen are combined to form HCl, if you are to obtain a 98% yield and you start with 3.0 moles of hydrogen, what amount of HCl is produced? Cl 2 + H 2  2HCl 3.0mol 98% = A (220)98% = A (220) A= (220)(.98) T 220 A=215.6gHCl What is the % yield of HCl if you start with 5.0 moles of hydrogen and obtain 345 grams of HCl? 345g x 100 = 345 x 100 = 95.8% T 360 ?g HCl = 5 mol H 2 x 2 mol HCl x g HCl = 364.6g HCl 11 mol H 2 1 mol HCl 360 g HCl (sigfigs)

22 LIMITING REAGENT LIMITING REACTANT We have already seen that reaction conditions are not ideal limit the amount of product produced when dealing with % yield calculations. Now suppose one reactant is completely consumed during the reaction? What is a limiting reactant? The reactant that is completely consumed during a reaction. Limits the amount of product formed. Excess reactant? The reactant that is left over after a reaction. The amount used depends on the limiting reactant.

23 Steps for determining the LR and EXCESS reagent 1. Set up 2 stoichiometry problems beginning with the given amount of each reactant and determine the amount of 1 product that may be formed. 2. Select the reagent that limits the amount of product by choosing the one that produces less product. This is your limiting reagent. 3. Determine how much of the other reactant is used starting with the limiting reagent and using stoichiometry. 4. Subtract the amount of excess reactant used from the given amount to find the amount left over.

24 LIMITING REAGENT example Zinc and sulfur react to form zinc sulfide. If grams react with grams of sulfur determine the following: 8Zn + S 8  8ZnS a. What is the limiting reagent? ?gZnS = g Zn x 1 mol Zn x 8 mol ZnS x 97.48g ZnS = g ZnS g Zn 8 mol Zn 1 mol ZnS ?gZnS = g S 8 x 1 mol S 8 x 8 mol ZnS x 97.48g ZnS = g ZnS g S 8 1 mol S 8 1 mol ZnS

25 b. Which reactant is in excess? S 8 c. How much zinc sulfide could be produced? g Zn d. What is the maximum amount of product? g ZnS e. How many grams of excess reagent (S 8 ) remain? ?gS 8 = g Zn x 1 mol Zn x 1 mol S 8 x g S 8 = 66.52g S 8 used used g Zn 8 mol Zn 1 mol S g S 8 given g S 8 used g S 8 left over


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