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SKILLS Project Balancing Chemical Reactions. What do you mean, balancing? According to the Law of Conservation of Mass, “matter is neither created nor.

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Presentation on theme: "SKILLS Project Balancing Chemical Reactions. What do you mean, balancing? According to the Law of Conservation of Mass, “matter is neither created nor."— Presentation transcript:

1 SKILLS Project Balancing Chemical Reactions

2 What do you mean, balancing? According to the Law of Conservation of Mass, “matter is neither created nor destroyed by normal chemical reactions.” In other words, “what goes in must come out.” A chemical reaction shows how chemicals interact to form new, different substances, but is subject to the laws of conservation as much as any other change- whether physical or chemical. Balancing is the process by which chemical reactions are corrected so that the reactants contain the same amount of matter as the products.

3 So how do we balance? As you know, a chemical reaction shows the relationships between moles of different substances. We create “balance” by adjusting the coefficients (numbers placed in front of substances in a reaction to represent moles). Typically, you start with a metal and work your way through the problem “back and forth.” Save hydrogen and oxygen for last. Balance hydrogen before oxygen.

4 Tips Be patient, you may have to re-work through a reaction more than once to balance everything. Pick a starting point (a metal usually works well) and work steadily from element to element. If, by balancing one element, you change another, you should balance that one next. Keep at it! Balancing takes time and you will pick up many shortcuts and tricks along the way.

5  As you know, we need to make these numbers equal to one another to satisfy the Law of Conservation of Mass. “What goes in, must come out!”  First, determine how many moles of this element, Fe, entered the reaction as reactants and then how many exited as products.  We have two choices, Al and Fe. Al is already balanced, so we’ll begin with Fe.  To correct this, we’ll add the coefficient “3” in front of the iron symbol in the products.  The coefficient, 3, multiplies the symbols directly behind it, and creates the 3 moles of Fe we need to balance that element.  We’re done with the Fe, so lets move on to N. We’re not going to touch Al yet because its still balanced.  Notice that 2 moles of nitrogen have entered the reaction, while only 1 has exited.  We’ll correct this by multiplying the AlN by 2. Note that this affects both the Al and the N.  To balance Al, we need to multiply the Al on the reactant side by 2, this creates two moles in both the reactants and products.  The last element that needs to be balanced is Al. By correcting the N, we disrupted the amount of Al as well. Now, 1 mole of Al enters as a reactant and two moles exit.  Now that we’ve balanced all 3 elements, we need to double check our work. Doing so, we find that Al = 2 mol, Fe = 3 mol, and N = 2 mol for both the reactants and products.  Everything checks out. All that is left is to place a “1” in the spaces we didn’t need to change. In this case, we didn’t need to adjust the Fe 3 N 2, so we can simply assume 1 mol is present.  Congratulations, you’ve balanced your first chemical reaction. You will note that this was a single replacement reaction. 1 mol 3 mol Example 1: Simple Balancing ___ Al + ___ Fe 3 N 2  ___ AlN + ___ Fe 3 Typically, metals make good starting points for balancing chemical reactions. Reactants:Products: Fe 3 mol Work steadily through the reaction, element to element. 2 N 2 mol 1 mol 2 mol 1 mol 2 mol Al 21 Go back and double-check your work. Everything will balance if you are truly finished.

6 Example 2: Simple Balancing ___ NaClO 3  ___ NaCl + ___ O 2 2 Typically, metals make good starting points for balancing chemical reactions. Reactants:Products: O Work steadily through the reaction, element to element.  As before, we’ll want to try to start balancing this reaction using a metal such as Na. However, it is already balanced as is the Cl. So, we have no choice but to start with oxygen. 3 mol 2 mol  Looking at the equation, we find that 3 mol of oxygen entered the reaction and 2 exited. However, we can’t turn a 3 into a 2 or vice versa- both will have to be multiplied.  For these situations, we use the least common multiple (LCM). Essentially this is the smallest number that both coefficients can be multiplied to. For 2 and 3, this number is 6. 3  By using the LCM, we’ve managed to balance oxygen on both sides of the reaction. However, this has disrupted both Na and Cl. 6 mol  You can balance either Na or Cl at this point, but I usually stick to correcting the metals first. This tends to save time and effort later on. Here, we should correct Na first.  Balancing the oxygens un-balanced the moles of Na. We now have 2 moles entering the reaction and 1 exiting. 2 mol 1 mol Na  As you learned in the first example, multiplying the NaCl on the products side by 2 will correct the moles of sodium quite easily. Note that this also affects the moles of chlorine, Cl. 2 mol 2  Our last element, Cl, must now be checked. Looking at what we have, it appears the chlorine has already been balanced with 2 moles entering and 2 exiting the reaction. 2 mol Go back and double-check your work. Everything will balance if you are truly finished.  After double-checking our balancing, we find that 2 mol of Na, 2 mol of Cl, and 6 mol of O all enter and exit the equation. All coefficients have been added, so no 1’s this time.  Good work, you’ve sucessfully balanced this decomposition reaction.

7 Example 3: Simple Balancing ___ C 3 H 8 + ___ O 2  ___ CO 2 + ___ H 2 O Typically, metals make good starting points for balancing chemical reactions. Reactants:Products: C Work steadily through the reaction, element to element. 3  This time around, we don’t have any metals to start us off. However, we know not to start with either hydrogen or oxygen, so we’ll begin with carbon, C. 3 mol 1 mol Go back and double-check your work. Everything will balance if you are truly finished.  Looking at the equation, we notice that 3 mol C enter the reaction and 1 exits.  We’ll correct this by multiplying carbon dioxide, CO 2, by a coefficient of 3. 3 mol  Next up, we have to choose between oxygen, O, and hydrogen, H. As a rule, we’ll want to tackle the hydrogen first. In most equations, oxygen is best saved for last. H  The equation shows us that 8 mol H entered and that only 2 exited (in H 2 O). 8 mol 2 mol  We’ll balance the hydrogens by multiplying the H 2 O by 4. This will balance out the product side of the reaction. 4 8 mol  Lastly, we’ll want to deal with the oxygen. Note that oxygen is contained in two different substances on the products side, H 2 O and CO 2.  Two moles of oxygen enter the reaction from O 2, while a total of 10 moles of oxygen exit. (6 moles oxygen from the 3CO 2 and 4 moles oxygen from the 4H 2 O.) 2 mol 10 mol  We can correct this by multiplying the O 2 in the reactants by a coefficient of mol  Upon reviewing the entire reaction, we find that 3 moles of carbon, 8 moles of hydrogen, and 10 moles of oxygen enter and exit the reaction. This agrees with the Law of Conservation.  We have completely balanced this combustion reaction.  As usual, we will place a “1” in front of C 3 H 8 as we haven’t needed to change it. 1

8 SO 3  In turn, this correction has disrupted potassium, K, so we’ll do that next.  This time around, we have both K and Al to choose from. Either one would be acceptable, but we’ll start with aluminum.  In order to balance out aluminum, we’ll need to multiply AlCl 3 by 2  Our corrections to the AlCl 3 have disrupted the Cl, so we’ll tackle it next.  We now have 6 mol Cl entering the equation as reactants and 1 mol Cl exiting as a product.  To correct this, multiply the KCl with a coefficient of 6, bringing both totals to 6 mol Cl. Example 4: Moderate Difficulty ___ AlCl 3 + ___ K 2 SO 3  ___ Al 2 (SO 3 ) 3 + ___ KCl Typically, metals make good starting points for balancing chemical reactions. Reactants:Products: Al Work steadily through the reaction, element to element. 2 1 mol 2 mol Go back and double-check your work. Everything will balance if you are truly finished.  1 mol Al enters the equation while 2 are exiting in Al 2 (SO 3 ) 3. 2 mol Cl 6 mol 1 mol 6 6 mol K  From the equation we see that 2 mol K are entering and 6 mol K are exiting. 2 mol 6 mol  Multiplying K 2 SO 3 with a coefficient of 3 will bring both totals to 6 mol K. 6 mol 3  Now, we are left with sulfur and oxygen. You will notice that SO 3 is a polyatomic and that it appears on both sides of the reaction intact. We can balance the entire thing together.  Notice that K 2 SO 3 can be written as K 2 (SO 3 ). So, 3K 2 (SO 3 ) produces 3 mol of SO 3 2- as does the Al 2 (SO 3 ) 3. A quick re- check reveals that both contain 3 mol S and 9 mol O as well. 3 mol  Learning to balance polyatomics, rather than breaking them into component elements, will save you a great deal of time and effort.  A quick re-check of our work reveals that 2 mol Al, 6 mol Cl, 6 mol K and 3 mol (SO 3 ) 2- [ or 3 mol S and 9 mol O] enter and exit the reaction. We finish by adding a 1 to Al 2 (SO 3 ) 3. 1  Although slightly more complicated, you have successfully balanced this double replacement reaction.

9  Now that Ca is corrected, we can tackle the nearby carbonate ion, CO 3 (2-). Example 5: Moderate Difficulty ___ CaCO 3 + ___ H 3 PO 4  ___ Ca 3 (PO 4 ) 2 + ___ H 2 CO 3 Typically, metals make good starting points for balancing chemical reactions. Reactants:Products: Ca Work steadily through the reaction, element to element. 3  To begin this problem, we only have one metal to choose from. We’ll start with calcium, Ca. 1 mol 3 mol Go back and double-check your work. Everything will balance if you are truly finished.  Looking at the equation, we see that 1 mol Ca enters and 3 mol Ca exit.  To correct this, we’ll multiply CaCO 3 by a factor of 3. CO 3 (2-)  3CaCO 3 produces 3 mol of carbonate entering the reaction, while H 2 CO 3 only contains 1 mol exiting. 3 mol 1 mol  We can correct this by multiplying H 2 CO 3 using a coefficient of 3, creating 3 mol of carbonate, CO 3 2-, on both sides. 3 3 mol  Next, we’ll want to skip hydrogen and tackle the PO 4 (3-) polyatomics (phosphate). PO 4 (3-)  From the equation, we can see that 1 mol of phosphate enters the equation while 2 mol exit. 1 mol 2 mol  To correct this, we need to multiply H 3 PO 4 by 2. This will create the 2 mol of PO 4 (3-) we need. 2 2 mol  Finally, we will deal with the hydrogen atoms. It is worth noting that many elements are already balanced by the time we get to the end, especially when balancing polyatomic ions. H  Reviewing the equation, we see that H is already balanced, with 6 moles entering as reactants in the 2H 3 PO 4 and exiting in the 3H 2 CO 3.. This is a good sign and indicates accurate work. 6 mol  Finally, we review our work and find that we have 3 mol Ca, 3 mol CO 3 (2-), 6 mol H, and 2 mol PO 4 (3-) entering as reactants and exiting as products. Place a 1 in front of Ca 3 (PO 4 ) 2. 1  Congratulations! This double replacement reaction involved several polyatomics and reflects the difficulty level that most chemical reactions tend to offer.

10  We’ve done every other element, so we should double check oxygen. Example 6: Moderate Difficulty ___ H 2 C 2 O 4 + ___ NaOH  ___ Na 2 C 2 O 4 + ___ H 2 O Typically, metals make good starting points for balancing chemical reactions. Reactants:Products: Na Work steadily through the reaction, element to element. 2  Once again, you only have one metal to choose from: Na. 1 mol 2 mol Go back and double-check your work. Everything will balance if you are truly finished.  From the equation, only 1 mol Na enters the reaction, while two exit in Na 2 C 2 O 4.  To correct this, we need to multiply the NaOH by 2. 2 mol  Next, we’ll target the polyatomic C 2 O 4 (2-) C 2 O 4 (2-)  However, we find C 2 O 4 (2-) to already be balanced with 1 mol on either side, so we can move on to hydrogen, H. 1 mol  Remember, we skip oxygen and save it for last in virtually all chemical reactions. H  Upon totaling up hydrogen, we find 4 mol H from H 2 C 2 O 4 and 2NaOH on the reactant side. The products contain 2 mol H in H 2 O.  To correct this, we will multiply the H 2 O in the products by a coefficient of 2. 4 mol 2 mol 4 mol 2  As we’ve already balanced the C 2 O 4 polyatomic (oxalate ion), we should really only count other oxygens. In this case, there are 2 on each side. Including those in oxalate, 6 mol O. O  Reviewing the reaction, there is a total of 4 mol H, 2 mol C, 6 mol O, and 2 mol Na in both the reactants and in the products. This equation is fully balanced. 2/6 mol  To finish balancing the reaction, we need to place “1” in front of H 2 C 2 O 4 and Na 2 C 2 O  Congratulations! This double replacement reaction was a bit trickier than others, but patience and careful work are usually all that is needed.

11 ___ C 8 H 18 + ___ O 2  ___ CO 2 + ___ H 2 O Example 7: High Difficulty Typically, metals make good starting points for balancing chemical reactions. Reactants:Products: C Work steadily through the reaction, element to element.  This example of a combustion reaction is very similar to example 3. We’ll need to start with carbon as there are no metals in this reaction. 8 mol 1 mol Go back and double-check your work. Everything will balance if you are truly finished.  From the equation, we see that 8 mol C enter as reactants while 1 mol C exits as a product, CO 2.  We’ll need to multiply CO 2 with a coefficient of 8. This will create 8 mol C in the products as well as reactants. 8 mol  Next up, we’ll tackle hydrogen. H  We have 18 mol H in the reactants through C 8 H 18 and 2 mol H in the products through H 2 O. 18 mol 2 mol  To correct this discrepancy, we’ll need to multiply the H 2 O by a factor of mol  Last, but not least, we need to tackle the oxygen. O  From the equation, 2 mol O enter the reaction from O 2 and 25 mol O exit through 8CO 2 and 9 H 2 O. We’ll have to use the LCM to balance these. The LCM of 2 and 25 is 50.  This poses a particular problem. We can multiply O 2 by 25 to create 50 mol O, but what about 8CO 2 and 9H 2 O? Together, they produce 25, so multiply both by 2 to create mol 25 mol 50 mol  Although we’ve corrected oxygen, we’ve doubled carbon to 16 mol and hydrogen to 36 mol. We’ll need to compensate for both of these by doubling C 8 H  Now, to check our results. Overall, there are 16 mol C, 36 mol H, and 50 mol O in both the reactants and in the products. This reaction is fully balanced.  Congratulations! Reactions like this are fairly difficult and can be exceedingly challenging. Practicing these reactions provides insight and skill into the art of balancing reactions.

12  Now that we’ve balanced everything: 6 mol V, 36 mol NO 3 1-, 9 mol Ag, and 12 mol N enter and exit this equation. Everything is properly balanced.  There are 3 mol Ag entering the reaction and 1 mol exiting in Ag(NO 3 ) 4.  To correct this, we’ll multiply the Ag(NO 3 ) 4 by a factor of 3. This will create the 3 mol of silver we need on both sides.  Next, we’ll need to correct our metals. Adjusting the NO 3 1- amounts affected both the V and Ag amounts.  Congratulations! This was a particularly difficult problem and, if you understand the previous steps, you are ready to tackle just about any problem you’ll face in this class.  To fix this, we’ll multiply the V(NO 3 ) 6 by a coefficient of 3. This brings the number of moles of vanadium to 3 on both sides of the reaction.  Next, tackle the nitrate (NO 3 ) 1- polyatomics. We’re going to do this first because, by adjusting the V and Ag amounts, we’ve also affected the NO 3 1- amounts.  Using the equation, 3V(NO 3 ) 6 = 18 mol NO 3 1- and 3Ag(NO 3 ) 4 = 12 mol NO  Notice that, for vanadium, we have 6 mol entering and 3 mol exiting. We’ll correct this by multiplying V 3 N 6 by 2.  Now, for Ag, we have 3 mol entering and 9 mol exiting. We’ll need to multiply Ag 3 N 4 by 3 to correct this imbalance.  Finally, we need to check out nitrogen, N.  By balancing everything else, we’ve also balanced out the nitrogen with 12 mol entering and 12 mol exiting. Example 8: High Difficulty ___ V(NO 3 ) 6 + ___ Ag 3 N 4  ___ V 3 N 6 + ___ Ag(NO 3 ) 4 Typically, metals make good starting points for balancing chemical reactions. Reactants:Products: V Work steadily through the reaction, element to element. 3  We’ll start out with V, vanadium. Technically, we could start out with Ag, silver, if you would like. 1 mol 3 mol Go back and double-check your work. Everything will balance if you are truly finished.  Using the equation, only 1 mol vanadium enters as a reactant while 3 exits as a product. 3 mol  Next, we’ll target the silver, Ag. 3 Ag NO 3 (1-) 69 3 mol 1 mol 2 3 mol  The LCM shared between 12 and 18 is 36. So, we’ll have to multiply the 3V(NO 3 ) 6 by 2 (3x2=6) and 3Ag(NO 3 ) 4 by 3 (3x3=9). So, 6V(NO 3 ) 6 and 9Ag(NO 3 ) mol 12 mol 36 mol V 6 mol 3 mol 6 mol3 mol 9 mol 3 AgN 12 mol

13 Practice On Your Own (I) ___ RbOH + ___ H 2 SO 4  ___ H 2 O + ___ Rb 2 SO 4 ___ C 4 H 10 + ___ O 2  ___ CO 2 + ___ H 2 O ___ CaCO 3  ___ CaO + ___ CO 2 ___ CO 2 + ___ H 2 O  ___ C 6 H 12 O 6 + ___ O 2 ___ C 8 H 18 + ___ O 2  ___ CO 2 + ___ H 2 O

14 ___ H 2 + ____ O 2  ___ H 2 O ___ Pb(NO 2 ) 2 + ___ NaCl  ___ PbCl 2 + ___ NaNO 2 ___ KClO 3  ___ KCl + ___ O 2 ___ Fe + ___ HCl  ___ H 2 + ___ FeCl 3 ___ C 2 H 2 + ___ Cl 2  ___ HCl + ___ C Practice On Your Own (II)

15 Congratulations! Balancing chemical equations is one of the most important skills you will learn this year. Master this skill and remember it well- many of the techniques and skills that follow are based on the successful construction of a fully balanced chemical reaction.


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