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Ch. 12 Stoichiometry AKA…… Chemistry Math….., Ooohhhh Scary!!

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Presentation on theme: "Ch. 12 Stoichiometry AKA…… Chemistry Math….., Ooohhhh Scary!!"— Presentation transcript:

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2 Ch. 12 Stoichiometry AKA…… Chemistry Math….., Ooohhhh Scary!!

3 I. The Arithmetic of Equations A. Using Everyday Equations 1. Equations are like recipes that tell chemists the amount of reactants to mix and products to expect. 2. The calculation of quantities in chemical reactions is called stoichiometry. This is a form of book keeping for reactions.

4 I. The Arithmetic of Equations B. Interpreting Chemical Equations 1.The information that can be derived from an equation a. the types of particle (formulas) b. number of moles c. mass (law of conservation of mass) d. volume ( only if at STP) 2. You can use this to solve several types of problems. Write the balanced equation. Then determine the mole to mole ratio.

5 What is Stoichiometry? Stoichiometry is a powerful tool that all chemist must utilize. It separates the scientists from everyone else. It allows a chemist to determine an amount of another compound that is needed or produced in a chemical reaction. Here is a four step process for solving problems.

6 How Do We Use It? Question- How many grams of water can be produced from 10.grams of H 2 with excess O 2 ? excess=you have more than enough O 2 to complete this reaction, worry about H 2 1. You need a balanced chemical reaction. –2H 2 + O 2 ==> 2H 2 O –(remember that the coefficients are really mole ratios)

7 How Do We Use It? 2. FIND THE MOLES of your compound. Sometimes it is already given, sometimes you have to calculate it. –Given==> 10. grams of H 2 –Find the moles=> 10.g H 2 (1mol / 2.0g H 2 )= 5.0 mol H 2

8 How Do We Use It? 3. DO THE RATIO-Now you can find the moles of any other compound in this reaction using the coefficients of the balanced reaction. Here are a few ways to do this. –Factor Label 5.0 mol H 2 ( 2 mol H 2 O /2 mol H 2 ) =5.0 mol H 2 O –or set up a proportion (5.0 mol H 2 / 2 H 2 )= (x mol H 2 O / 2 H 2 O ) x= 5.0 mol H 2 O –or set up an equation proportion –(remember that the coefficients are really mole ratios) 5.0 mol H 2 /2H 2 + ? mol O 2 / O 2 ===> ? mol H 2 O/2H 2 O Now you solved for O 2 also(2.5 mol) and H 2 O (5.0 mol)

9 How Do We Use It? 4. ANSWER THE QUESTION-Now find the mass from the number of moles –5.0 mol H 2 O ( 18.0g/mol)= 90. g H 2 O –If I wanted to know about the O 2, I could just take the moles and find the mass. –2.5 mol O 2 ( 32.0g/mol)= 80. g O 2 – FYI the mass in equals the mass out. –10.g/ 2H g/ + O 2 ===> 90.g /2H 2 O Conclusion- Find the moles, do the ratio, answer the question.

10 1. Write the balanced equation. 2. Find the number of moles for the given substance. 3. Look at the balanced equation to determine the ratio of moles of required substance to moles of given substance. 4. Convert the number of moles of required substance to grams. Method 2 - Here is another method that puts it all together in one equation.

11 If Iron (III) Sulfide, Fe 2 S 3 is removed from coal, it reacts with the oxygen in the air. How much Sulfur dioxide is formed from 100 g Fe 2 S 3 ? 1. Write a balance equations 2Fe 2 S 3 + 9O > 2Fe 2 O 3 +6SO 2 2. Write down the mass given and unknown. 100 g ? 2Fe 2 S 3 + 9O > 2Fe 2 O 3 +6SO 2

12 3.Calculate the number of moles and use the factor-label method to determine the mass or volume of your unknown. (100gFe 2 S 3 )1 moleFe 2 S 3 6moleSO 2 64gSO 2 160gFe 2 S 3 2moleFe 2 S 3 1moleSO 2 =120g SO 2

13 What is the mass of Oxygen produced by the decomposition of 30g of Potassium Chlorate? 30g ? 2KClO > 2KCl +3O 2 30 gKClO 3 1moleKClO 3 3moleO 2 32g O gKClO 3 2mole KClO 3 1moleO 2 = 11.7 g O 2

14 II. Chemical Calculation B. Solving Other Stoichiometric Problems 1. Write the balanced equation. 2. Find the number of moles for the given substance. 3. Compare the mole to mole ratio. You can then solve for mass, volume, or number of particles.

15 III. Limiting Reagent and Percent Yield A. What is a Limiting Reagent? 1. A limiting reagent determine the amount of product that can be produced in a reaction. 2. The limiting reagent is the reactant that will be use up first in the reaction. 3. When solving a limiting reagent problem, the first step is to convert the reactants to moles. By comparing the number of moles, you sometimes can determine the limiting reagent.

16 III. Limiting Reagent and Percent Yield B. Percent Yield 1. Percent yield is the comparison of the actual amount of product by the amount of product calculated. 2. The formula for percent yield is the actual amount of product divided by the theoretical amount of product multiplied by 100. Percent Yield = actual amount of product x 100 theoretical amount of product

17 Percent Yield Calculations Percent yield calculations differ from limiting stoichiometry problems by only one extra step. The question includes a mass recovered. Here is the formula you will use at the end of the problem. When you read the question the mass of the product must not be used until the very end. So.....put it in your pocket for later. You are allowed to use moles, grams, liters in this equation, as long as actual and theoretical are both in these units.

18 EXAMPLE QUESTION #1 (not limiting) 2CO (g) + O 2(g) --> 2CO 2(g) Calculate the % yield if 69.1g of CO combines with excess O 2 to form an experimental yield of 48.3L of CO –See how it says yields 48.3L of CO 2 save that for later (in your pocket). –Now determine the volume of of CO 2 from 69.1g of CO using stoichiometry

19 Step 2 Find the moles, the mol ratio, then find the volume of CO 2 –69.1g CO(1 mol CO/28.0g)= 2.57mol CO –2.57mol CO(2mol CO 2 /2mol CO)= 2.57mol CO 2 –2.57mol CO 2 (22.4L/1 mol)= 55.3L of CO 2 –(69.1 g CO)(1 mol CO/28.0g)(2mol CO 2 /2mol CO)(22.4L/1 mol) = – = 55.3 L CO 2 Now take out that volume recovered and plug into the % yield equation. –(48.3L/55.3L) x 100%= 77.3% yield

20 Initial-Change-End Box Method Here are the Rules from the other Page –STEP1= SET UP the ICE Box –STEP 2- Find the moles, This is where you have to problem solve. –STEP 3- Find X, find the moles of everything –STEP 4- Answer the questions, convert moles to mass The question- –You have 20.0 g of elemental sulfur, S, and grams of O 2. What mass of SO 2 can be formed? How much reactant is left over? –S (s) + O 2(g) ==> SO 2(g)

21 STEP1= SET UP the ICE Box below the balanced reaction. (I-Initial C-Change E-End). S (s) +O 2(g) ==> SO 2(g) I-Initial C-Change-X +X E-End Using the coefficients, (the #'s in front of each compound) to determine the relative change. If there is a 2 in front, use 2X, if there is a 3 use 3X. Also reactants decrease in their amounts (-X), products will increase (+X).

22 Step 2- Moles are the only thing allowed in the ice box. Find the moles of each compound and insert that value in its “I” box. Moles S –20.0g S X 1 mole/ 32.1g –=0.623 moles S Moles O 2 –160.0g O 2 X 1 mole/32.0g –=5.000 moles O 2 S (s) +O 2(g) ==> SO 2(g) I-Initial C-Change-X +X E-End

23 Step 3- Find X, one of the reactants is limiting, which means it runs out. You end up with 2 possible scenarios for this reaction. –if S runs out ==> mol -X =O ; X is therefore mol –if O 2 runs out ==> mol -X=O ; X is therefore mol Which ever reactant gives you the lower value for X is the limiting reactant and this X value is applied as X in your ICE BOX.

24 Solve the Problem (If you applied the incorrect, larger X value, will get a negative amount at the end, so go back and change it.) Now solve for everything (add or subtract down each column). S (s) +O 2(g) ==> SO 2(g) I-Initial C-Change E-End

25 Step 4- Answer the questions, you have them moles of everything. Convert to grams. What mass of SO 2 can be formed? –0.623mol SO 2 X 64.1g/1 mol –=39.9g SO 2 How much reactant is left over? –4.377mol O 2 X 32.0g/ 1 mol –=140.1g SO 2

26 Level 2- The equation now has coefficients. If you are provided 200.g of sodium and 250. grams of iron (III) oxide, which substance is the limiting reactant? How many grams of Iron are produced? How much reactant is in excess? –6Na + Fe 2 O 3 --> 3Na 2 O + 2Fe

27 STEP1= SET UP the ICE Box below the balanced reaction. (I-Initial C-Change E-End). Using the coefficients, (the #'s in front of each compound) to determine the relative change. If there is a 2 in front, use 2X, if there is a 3 use 3X. Also reactants decrease in their amounts (-X), products will increase (+X). 6Na +Fe 2 O 3 ==> 3Na 2 O + 2Fe I-Initial C-Change-6X-X+3X+ 2X End

28 Step 2- Moles are the only thing allowed in the ice box. Find the moles of each compound and insert that value in its I box. Moles Na –200.g Na X 1 mole/ 23.0g –=8.70 moles Na Moles Fe 2 O 3 –250.g Fe 2 O 3 X 1 mole/ 159.7g –=1.57 moles Fe 2 O 3 6Na +Fe 2 O 3 ==> 3Na 2 O + 2Fe I-Initial C-Change-6X-X+3X+2X End

29 Step 3- Find X, one of the reactants is limiting, which means it runs out. You end up with 2 possible scenarios for this reaction. if Na runs out ==> 8.70 mol -6X =O ; X is therefore 1.45 mol if Fe 2 O 3 runs out ==> 1.57mol -X=O ; X is therefore 1.57 mol 6Na +Fe 2 O 3 ==> 3Na 2 O + 2Fe I-Initial C-Change-6X-X+3X+2X End

30 Which ever reactant gives you the lower value for X is the limiting reactant and this X value is applied as X in your ICE BOX. Na is therefore limiting. –(If you applied the incorrect, larger X value, will get a negative amount at the end, so go back and change it.) –Now solve for everything (add or subtract down each column). 6Na +Fe 2 O 3 ==> 3Na 2 O + 2Fe I-Initial C-Change-6(1.45) (1.45)+2(1.45) End

31 Step 4- Answer the questions, you have them moles of everything. Convert to grams. How many grams of Iron are produced? –2.90 mol Fe X 55.85g/1 mol –=162g Fe How much reactant is in excess? –0.12 mol Fe 2 O 3 X 159.7g/1 mol –=19.2g Fe 2 O 3

32 Work Cited Textbook: Chemistry, Wilbraham, Staley, Matta, Waterman; Addison- Wesley


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