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UT CH 301 - Reaction Stoichiometry

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HIGH SCHOOL CHEMISTRY REVIEW LECTURE 2: REACTION STOICHIOMETRY Chapter summary. We just learned that simple quantitative relationships based upon the idea of the law of simple proportions could be combined with other concepts from Dalton’s Atomic Theory to create a host of problems based upon the quantitative relationships between atoms in molecules. We learned to use unit factors to convert between grams and moles moles and # of atom or molecules atomsA and moleculeA Now we take things just a single step further by applying these concepts of stoichiometry to CHEMICAL REACTIONS. After all, since we know there is conservation of mass in the world, then when a chemical change occurs and new compounds are formed by chemical reaction, there must be a collection of simple unit factors that describe these changes. The additional unit factor is simply asking for a stoichiomeric relationship between numbers of atoms or molecules of one chemical species and another chemical species involved in a reaction.Example. Consider the combustion of methane, CH4: methane + oxygen Æ water and carbon dioxide Writing it without concern for stoichiometry we have CH4 + O2 Æ CO2 + H2O (which is unbalanced and violates the conservation of matter) But by inspection we can find a way for conserve matter in a balanced reaction by placing a coefficient of 2 in front of the diatomic oxygen and in front of the water: CH4 + 2O2 Æ CO2 + 2H2O (balanced, matter is conserved, all is right in the world) From this balanced equation we can form all kinds of units factors relating molecules or atoms of reactants to molecules or atoms of products. For example there is: one mole of CH4 for every one mole of H2O there are: 2 molecules of O2 for every one molecule of CO2 there are even: four atoms of H in CH4 for every two atoms of O in 2H2Oand many more, many more unit factors. This means that we can do all the stoichiometry problems from compositional stoichiometry with the added twist that we can add unit factors for reactions. Balancing Chemical Equations One annoying feature of using chemical reactions is that sometimes the equations you are given are not balanced (i.e., they don’t satisfy the law of conservation of mass). For example the first methane combustion equation above was not balanced. Well you can’t use unit factors if don’t have balanced equations, so we need some practice with balancing. So let’s try practicing with another reaction. This time propane combustion. C3H8 + O2 Æ CO2 + H2O unbalanced Coefficients are 1, 1, 1, 1 = 4 total molecules in unbalanced reaction. C3H8 + O2 Æ 3CO2 + 4H2O unbalanced Coefficients are 1, 1, 3, 4 = 9 total molecules in unbalanced reaction. C3H8 + 5O2 Æ 3CO2 + 4H2O balanced Coefficients are 1, 5, 3, 4 = 13 total molecules in a balanced reaction. How did he do that? A few hints for balancing equations (these don’t work for complicated redox problems, but they are a way to start with simpler equations. 1. First balance the compounds which contain elements that appear only once on each side of the equation. For example, in the propane combustion reaction above, note that C only appears once on the left (in C3H8) and once on the right (in CO2) above. BALANCE IT FIRST. Note that oxygen atoms appear in two compounds on the right (CO2 and H2O), do it later.2. Save the elements or atoms with a single element to balance last (He, H2, N2, O2, P4, S8,....). For example, solve for O2 last in the propane problem above. If you follow these two rules for balancing equations, solving every equation will be a snap---unless the rules don’t apply. Then you are on your own and will wish you’d practiced more. Look, everyone can balance equations. It is just a matter of how much time it takes: practice a lot and solve your exams equation in 30 seconds, practice while taking the exam, and do the problem in five minutes. It is up to you. A little practice for right now. Try to follow the rules: Example 1. Balance NaCl + H2O Æ NaOH + Cl2 + H2 Start with Na that appears only once on either side and end with Cl2 or H2, which are elements: Step one, Na are already balanced: NaCl + H2O Æ NaOH + Cl2 + H2 Step two, two Cl on right means need to add two on left: 2NaCl + H2O Æ NaOH + Cl2 + H2 Step 3, oops, now Na are not balanced, need to add two to right side 2NaCl + H2O Æ 2NaOH + Cl2 + H2 Step 4, now balance the O: 2NaCl + 2H2O Æ 2NaOH + Cl2 + H2 Step 5, finally, balance the H--oh, it already is: 2NaCl + 2H2O Æ 2NaOH + Cl2 + H2 Example 2. Balance: H2SO4 + NH3 Æ (NH4)2SO4 By inspection, add a 2 in front of the ammonia H2SO4 + 2NH3 Æ (NH4)2SO4That was easy. Example 2. Balance by yourself. Ba(OH)2 + P4O10 -----> Ba3(PO4)2 + H2O That wasn’t so easy, was it? Here is the answer. 6Ba(OH)2 + P4O10 -----> 2Ba3(PO4)2 + 6H2O Okay, so the rules don’t always work. Actually balancing chemical reactions can get very difficult when oxidation numbers change during a reaction. We’ll explore more systematic ways to balance chemical reactions toward the end of CH302 when we start learning about electrochemistry. For now, assume that simple “inspection” will do the trick most of the time. Reaction Stoichimetry Calculations Assuming that you have a balanced equation, you can do a whole host of problems that involve converting between grams of this and atoms of that which are conceptually identical to compositional stoichiometry problems. For example, consider how many units factors there are for C3H8 + 5O2 -----> 3CO2 + 4H2O There are a bunch, like: • 1 molecule of propane for every 3 molecules of CO2 • 3 atoms of C in propane for 4 molecules of water • 9 atoms of hydrogen in propane for 8 atoms of hydrogen in water and on and on? How many all together? Would you believe hundreds?Now on to some problems. For the following problems, use the reaction Fe2O3 + 3CO ----> 2Fe + 3CO2 Problem 1. How many CO molecules are required to react with 25 formula units of Fe2O3? ? CO molecules = (25 Fe form. unit3


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UT CH 301 - Reaction Stoichiometry

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