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Stoichiometry

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Bellringer Write the quantities of ingredients you would use to make a bologna and cheese sandwich. Then determine how many sandwiches you could make from 24 slices of bread. Calculate how much of each ingredient is needed.

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**Balanced Equations Show Proportions**

Coefficients show the proportions of the reactants and products. 2H2 + O2 2H2O Calculations involving chemical reactions use these proportions to find amounts of reactants and products. Use models to demonstrate this reaction and show the molecules. Tell students to assume: **For each problem in this section, assume that there is more than enough of all other reactants to completely react with the reactant given. **Also assume that every reaction happens perfectly, so that no product is lost during collection.

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**Relative amounts in equations can be expressed in moles**

Coefficients represent the moles of each substance. 2H2 + O2 2H2O 2 mol H2 +1 mol O2 form 2 mol H2O.

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**Stoichiometry You can determine:**

amount of reactant needed, or amount of product formed This proportional relationship between chemical coefficients is called stoichiometry.

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**OH, CHEMISTRY Verse 1 We started something new called stoichiometry.**

(THE STOICHIOMETRY SONG) Sung to the tune of "Jingle Bells.“ Lyrics by: Brandi Abbott 1992 Verse We started something new called stoichiometry. It's such a funny word. It brings a smile to me. The first step’s not so hard. It's something that is old. You take the grams that's given there and convert them into moles.

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**You have to really know your stuff for stoichiometry!**

Chorus Ooooooh Chemistry, chemistry, How exciting it can be! You have to really know your stuff for stoichiometry! (REPEAT)

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**It really doesn’t boast. You put the moles in a ratio **

Verse The second step's not big. It really doesn’t boast. You put the moles in a ratio To show least over most. When grams are what you seek a third step must be used. The least known's weight per mole will provide the final clue!

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**Always, the same 3 steps Grams Moles Molar Ratio**

Coeff. of what you want Coeff. of what you have Only if ans. is to be in grams Molecular weight of what you want 1 mole

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Mass- Mass What mass of hydrogen will be used when 14 g of nitrogen gas is reacted to produce ammonia gas? N H2 2NH3 14 g x 14 g N2 1mol N mol H g H = 3 g H2 28 g N mol N mol H2 grams to moles least/most least known's weight per mol

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Mole-Mass How many grams of sodium will be required to react with sufficient chlorine in order to produce 4.50 moles of NaCl? 2Na Cl2 2NaCl x mol 4.50 mol NaCl 2 mol Na g Na = g Na 2 mol NaCl 1 mol Na moles least/most least known's weight (given) per mol

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**Moles to Moles 2H2 + O2 2H2O 6.00 mol H2O 2 mol H2 = 6.00 mol H2**

6.00 moles of water vapor are required to assist oxygen delivery in surgical ventilators. How many moles of hydrogen and how many moles of oxygen must be introduced from the gas tanks in order to insure this amount? 2H2 + O2 2H2O x y mol 6.00 mol H2O 2 mol H2 = 6.00 mol H2 2 mol H2O moles least/most 6.00 mol H2O 1 mol O2 = 3.00 mol O2

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**Stoichiometry practice**

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**Limiting Reactants According to the periodic table 2Na + Cl2 2NaCl**

46 g g g Reactions do not always react perfectly. Temperature Age Humidity 100% of expected product rarely achieved.

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**Major PROBLEM in Industry**

Companies must produce enough product to supply their customers. They MUST use enough reactants to fill their orders WITHOUT producing excessive overage stored ($ lost), or thrown away ($ lost)

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**Industry adds an EXCESS of the cheapest reactant.**

Reactant that is completely consumed (used up) in the rxn. LIMITING REACTANT The limiting reactant’s value is used for all stoichiometry calculations.

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Too Much Info! How many moles of salt can be produced from the amounts of reactants given in the following equation? 2Na Cl2 2NaCl 46 g g ? Too much information one number gives reliable product information. one number lies!

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In our example 2Na Cl2 2NaCl 46 g g ? CALCULATE the amount of product from each amount… 46 g of sodium can produce 117 g of salt 91 g of chlorine can produce g of salt Therefore, Sodium is the limiting reactant and 117 g is the maximum amount of product that can be made.

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Check Both Numbers The calculation with the smaller answer is the correct product calculation and used the limiting reactant as its ‘have’ value.

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**How many grams of the excess reactant are ‘left over?’**

First determine how many grams of Cl2 were used. 2Na Cl2 2NaCl 46 g x 46 g Na 1 mol Na 1 mol Cl g Cl2 = 71 g Cl2 used 23 g Na 2 mol Na 1 mol Cl2 Subtract the amount used from the total amount present. [91 g present] - [71 g used] = 20 g Cl2 left over Θ

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Percent Yield The amount of product actually produced as compared to the possible (calculated) amount. % = Actual Amount or Actual Amount Possible Amount Stoich Answer

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Sample Calculation When 45.8 g of potassium carbonate are reacted completely w/ excess HCl, 46.3 g of potassium chloride are produced. Water and carbon dioxide are also produced. Calculate both the theoretical and % yield. K2CO HCl 2KCl H2O CO2 45.8 g x Theoretical (Stoich) 45.8 g K2CO3 1 mol K2CO3 2 mol KCl g KCl = g KCl 138 g K2CO3 1 mol K2CO3 1 mol KCl % Yield 46.3 g KCl x = % actual yield stated in question 49.1 g KCl

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**Mass - Energy Calculations**

Balanced equations can be used to determine the amt. of energy absorbed / released during the reaction. Treat energy as if it were a reactant or product. Heat of Reaction - qr If qr is (-) then products have less energy than the reactants - exothermic If qr is (+) then products have more energy than the reactants-endothermic.

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Sample Calculation Given: 2Na2O2 + 2H2O 4NaOH + O kJ How much E is released when 1.99 g of Na2O2 is reacted? 1.99 Na2O mol Kj = 2.75 Kj .78.0 g mol qr = negative, exothermic rxn qr = Kj

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Mathematics of Chemical Equations By using “mole to mole” conversions and balanced equations, we can calculate the exact amounts of substances that will.

Mathematics of Chemical Equations By using “mole to mole” conversions and balanced equations, we can calculate the exact amounts of substances that will.

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