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Stoichiometry Chemistry Chapter 12 " That which you persist in doing becomes easy to do - not that the nature of the thing has changed, but your power.

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Presentation on theme: "Stoichiometry Chemistry Chapter 12 " That which you persist in doing becomes easy to do - not that the nature of the thing has changed, but your power."— Presentation transcript:

1 Stoichiometry Chemistry Chapter 12 " That which you persist in doing becomes easy to do - not that the nature of the thing has changed, but your power and ability to do has increased." -- H.J. Grant

2 Chemistry & Baking  Chemistry is a lot like baking. What do you need to bake chocolate chip cookies? –Recipe  If you need more cookies, what are you going to do to the recipe?  In this example what are the reactants & the products? A balanced chemical equation provides the same kind of quantitative information that a recipe does!

3 Using Balanced Chemical Equations  Chemist use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction. –If you know the quantity of one substance in a reaction, you can calculate the quantity of any other substance consumed or created.  The calculation of chemical quantities in chemical reactions is known as stoichiometry.

4 Interpreting Chemical Equations  A balanced chemical equation can be, such as:  A balanced chemical equation can be interpreted in terms of different quantities, such as: –# of atoms - # and types of atoms –# of molecules - Coefficients of a balanced equation indicate the relative # of molecules –Moles – Coefficients of a balanced equation indicate the relative # of moles  Mass – can be calculated using mole / mass relationship; law of conservation of mass.  Volume – can be calculate using mole / volume relationship.

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6 From Balanced Equations to Moles AAAA balanced equation is essential for all calculations involving amounts of reactants & products. N2(g) + 3H2(g) -- 2NH3(g) TTTThe most important interpretation of this equation is that 1 mol of nitrogen reacts with 3 mol of hydrogen to form 2 mol of ammonia. –B–B–B–Based on this interpretation you can write ratios that relate moles of reactants to moles of product.

7 Mole Ratios  A derived from the coefficients of a balanced chemical equation  A mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation N 2 ( g ) + 3H 2 ( g ) --  2NH 3 ( g ) 1 mol N 2 3 mol of H 2 2 mol of NH 3 1 mol of N 2 3 mol of H 2 2 mol of NH 3

8 Using Mole Ratios  In chemical calculations, mole ratios are used to convert between –moles of reactant and moles of product –moles of reactants –moles of products. Mole ratio is a conversion factor

9 Steps to Solve Stoichiometric Calculations In a typical stoichiometric problem: Balance the equation Convert given quantity to moles (if needed) Then the mole ratio is used to convert from moles of given to moles of wanted Finally, the wanted moles are converted to specified unit of measurement, as required by the problem.

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14 More on Cooking & Chemistry  When baking, after getting your recipe, what do you do? –Check ingredients  What happens if the cookie recipe calls for 4 eggs and you only have 2? –Your recipe will be limited by the fact that you only have 2 eggs; the eggs are the limiting ingredients. A chemist often faces similar situations.

15 Limiting & Excess Reagents In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. LLLLimiting reagents – the reagent that determines the amount of product that is formed by a reaction. (makes the least) –T–T–T–The reaction occurs until the limiting reagent is used up! EEEExcess reagent – the reactant that is not used up.

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17 What do we have to do next???

18 Now that we have moles, what next??

19 Knows Limiting reagent is Cu at 1.26 mol (from previous problem) 1 mol of Cu 2 S has a molar mass of 159.1g Next ??? What is mole ratio of Cu to Cu 2 S??

20 Now we have everything we need to calculate how much product will be formed. So what’s next ?? We should have started with 80.0g of Cu, however, we had already converted the 80.0g to moles (in previous problem).

21 Percent Yield WWWWhen an equation is used to calculate the amount of product that will be formed during a reaction, the calculated value represents the theoretical yield or the m ax product that could be produced from given amounts of reactants. TTTThe amount of product actually formed during the reaction (in lab) is called the actual yield. TTTThe percent yield is a ratio of the actual to theoretical yield expressed as a percent.

22 Percent Yield (cont) The percent yield is a measure of the efficiency of a reaction carried out in the laboratory.

23 So how are we going to solve it??

24 What is mole ratio of CaCO 3 to CaO ?? We have everything we need so let’s solve it !

25 Now let’s calculate the percent yield. We have what we need so let’s solve it !


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