SKILLS Project

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. Example 1: Simple Balancing 3 2 1 2 ___ Al + ___ Fe3N2 ___ AlN + ___ Fe 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. 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. First, determine how many moles of this element, Fe, entered the reaction as reactants and then how many exited as products. Notice that 2 moles of nitrogen have entered the reaction, while only 1 has exited. We have two choices, Al and Fe. Al is already balanced, so we’ll begin with Fe. 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. 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!” 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 Fe3N2, so we can simply assume 1 mol is present. We’ll correct this by multiplying the AlN by 2. Note that this affects both the Al and the N. Congratulations, you’ve balanced your first chemical reaction. You will note that this was a single replacement reaction. 1 mol 3 mol 1 mol Reactants: 2 mol 2 mol Products: 1 mol 2 mol 3 mol 2 mol Fe Al N • Go back and double-check your work. Everything will balance if you are truly finished. • Work steadily through the reaction, element to element. • Typically, metals make good starting points for balancing chemical reactions.

6. Example 2: Simple Balancing 2 2 3 ___ NaClO3 ___ NaCl + ___ O2 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. 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. Balancing the oxygens un-balanced the moles of Na. We now have 2 moles entering the reaction and 1 exiting. 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. 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. By using the LCM, we’ve managed to balance oxygen on both sides of the reaction. However, this has disrupted both Na and Cl. 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. Good work, you’ve sucessfully balanced this decomposition reaction. 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. 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. 2 mol Reactants: 2 mol 2 mol 6 mol 3 mol Products: 1 mol 2 mol 6 mol 2 mol Na O • Work steadily through the reaction, element to element. • Go back and double-check your work. Everything will balance if you are truly finished. • Typically, metals make good starting points for balancing chemical reactions.

7. Example 3: Simple Balancing 5 3 4 1 ___ C3H8 + ___ O2 ___ CO2 + ___ H2O Looking at the equation, we notice that 3 mol C enter the reaction and 1 exits. 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. 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. As usual, we will place a “1” in front of C3H8 as we haven’t needed to change it. We’ll correct this by multiplying carbon dioxide, CO2, by a coefficient of 3. We have completely balanced this combustion reaction. We can correct this by multiplying the O2 in the reactants by a coefficient of 5. The equation shows us that 8 mol H entered and that only 2 exited (in H2O). We’ll balance the hydrogens by multiplying the H2O by 4. This will balance out the product side of the reaction. Lastly, we’ll want to deal with the oxygen. Note that oxygen is contained in two different substances on the products side, H2O and CO2. Two moles of oxygen enter the reaction from O2, while a total of 10 moles of oxygen exit. (6 moles oxygen from the 3CO2 and 4 moles oxygen from the 4H2O.) 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. 1 mol Reactants: 10 mol 2 mol 8 mol 3 mol Products: 10 mol 3 mol 8 mol 2 mol C H • Work steadily through the reaction, element to element. • Go back and double-check your work. Everything will balance if you are truly finished. • Typically, metals make good starting points for balancing chemical reactions.

8. Example 4: Moderate Difficulty 2 3 1 6 ___ AlCl3 + ___ K2SO3 ___ Al2(SO3)3 + ___ KCl 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 AlCl3 by 2 Our corrections to the AlCl3 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. Learning to balance polyatomics, rather than breaking them into component elements, will save you a great deal of time and effort. From the equation we see that 2 mol K are entering and 6 mol K are exiting. Notice that K2SO3 can be written as K2(SO3). So, 3K2(SO3) produces 3 mol of SO32- as does the Al2(SO3)3. A quick re-check reveals that both contain 3 mol S and 9 mol O as well. Now, we are left with sulfur and oxygen. You will notice that SO3 is a polyatomic and that it appears on both sides of the reaction intact. We can balance the entire thing together. 1 mol Al enters the equation while 2 are exiting in Al2(SO3)3. Multiplying K2SO3 with a coefficient of 3 will bring both totals to 6 mol K. Although slightly more complicated, you have successfully balanced this double replacement reaction. A quick re-check of our work reveals that 2 mol Al, 6 mol Cl, 6 mol K and 3 mol (SO3)2- [ or 3 mol S and 9 mol O] enter and exit the reaction. We finish by adding a 1 to Al2(SO3)3. 6 mol 6 mol 2 mol Reactants: 1 mol 6 mol 2 mol 2 mol 3 mol Products: 1 mol 6 mol 3 mol SO3 Cl Al K • Work steadily through the reaction, element to element. • Go back and double-check your work. Everything will balance if you are truly finished. • Typically, metals make good starting points for balancing chemical reactions.

9. Example 5: Moderate Difficulty 3 2 1 3 ___ CaCO3 + ___ H3PO4 ___ Ca3(PO4)2 + ___ H2CO3 Looking at the equation, we see that 1 mol Ca enters and 3 mol Ca exit. To correct this, we’ll multiply CaCO3 by a factor of 3. Congratulations! This double replacement reaction involved several polyatomics and reflects the difficulty level that most chemical reactions tend to offer. 3CaCO3 produces 3 mol of carbonate entering the reaction, while H2CO3 only contains 1 mol exiting. We can correct this by multiplying H2CO3 using a coefficient of 3, creating 3 mol of carbonate, CO32-, on both sides. Next, we’ll want to skip hydrogen and tackle the PO4(3-) polyatomics (phosphate). To begin this problem, we only have one metal to choose from. We’ll start with calcium, Ca. To correct this, we need to multiply H3PO4 by 2. This will create the 2 mol of PO4(3-) we need. 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. Reviewing the equation, we see that H is already balanced, with 6 moles entering as reactants in the 2H3PO4 and exiting in the 3H2CO3.. This is a good sign and indicates accurate work. Finally, we review our work and find that we have 3 mol Ca, 3 mol CO3(2-), 6 mol H, and 2 mol PO4(3-) entering as reactants and exiting as products. Place a 1 in front of Ca3(PO4)2. Now that Ca is corrected, we can tackle the nearby carbonate ion, CO3(2-). From the equation, we can see that 1 mol of phosphate enters the equation while 2 mol exit. 6 mol 3 mol Reactants: 2 mol 1 mol 1 mol 3 mol 6 mol Products: 2 mol 3 mol 1 mol Ca H • Work steadily through the reaction, element to element. • Go back and double-check your work. Everything will balance if you are truly finished. • Typically, metals make good starting points for balancing chemical reactions. CO3(2-) PO4(3-)

10. Example 6: Moderate Difficulty 1 2 1 2 ___ H2C2O4 + ___ NaOH  ___ Na2C2O4 + ___ H2O To correct this, we need to multiply the NaOH by 2. From the equation, only 1 mol Na enters the reaction, while two exit in Na2C2O4. Congratulations! This double replacement reaction was a bit trickier than others, but patience and careful work are usually all that is needed. However, we find C2O4(2-) to already be balanced with 1 mol on either side, so we can move on to hydrogen, H. Next, we’ll target the polyatomic C2O4(2-) Remember, we skip oxygen and save it for last in virtually all chemical reactions. Once again, you only have one metal to choose from: Na. To correct this, we will multiply the H2O in the products by a coefficient of 2. As we’ve already balanced the C2O4 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. 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. To finish balancing the reaction, we need to place “1” in front of H2C2O4 and Na2C2O4. We’ve done every other element, so we should double check oxygen. Upon totaling up hydrogen, we find 4 mol H from H2C2O4 and 2NaOH on the reactant side. The products contain 2 mol H in H2O. 2 mol Reactants: 2/6 mol 1 mol 4 mol 1 mol Products: 1 mol 2/6 mol 2 mol 2 mol 4 mol Na O H • Go back and double-check your work. Everything will balance if you are truly finished. • Work steadily through the reaction, element to element. • Typically, metals make good starting points for balancing chemical reactions. C2O4(2-)

11. Example 7: High Difficulty 2 25 16 8 18 9 ___ C8H18 + ___ O2 ___ CO2 + ___ H2O 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. We’ll need to multiply CO2 with a coefficient of 8. This will create 8 mol C in the products as well as reactants. 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. Next up, we’ll tackle hydrogen. 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. From the equation, we see that 8 mol C enter as reactants while 1 mol C exits as a product, CO2. To correct this discrepancy, we’ll need to multiply the H2O by a factor of 9. Last, but not least, we need to tackle the oxygen. From the equation, 2 mol O enter the reaction from O2 and 25 mol O exit through 8CO2 and 9 H2O. 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 O2 by 25 to create 50 mol O, but what about 8CO2 and 9H2O? Together, they produce 25, so multiply both by 2 to create 50. 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 C8H18. We have 18 mol H in the reactants through C8H18 and 2 mol H in the products through H2O. 1 mol 8 mol Reactants: 18 mol 50 mol 2 mol 8 mol Products: 18 mol 50 mol 25 mol 2 mol H O C • Go back and double-check your work. Everything will balance if you are truly finished. • Work steadily through the reaction, element to element. • Typically, metals make good starting points for balancing chemical reactions.

12. Example 8: High Difficulty 2 3 6 3 3 9 ___ V(NO3)6 + ___ Ag3N4 ___ V3N6 + ___ Ag(NO3)4 Using the equation, 3V(NO3)6 = 18 mol NO31- and 3Ag(NO3)4 = 12 mol NO31-. Next, tackle the nitrate (NO3)1- polyatomics. We’re going to do this first because, by adjusting the V and Ag amounts, we’ve also affected the NO31- amounts. To fix this, we’ll multiply the V(NO3)6 by a coefficient of 3. This brings the number of moles of vanadium to 3 on both sides of the reaction. Notice that, for vanadium, we have 6 mol entering and 3 mol exiting. We’ll correct this by multiplying V3N6 by 2. Now, for Ag, we have 3 mol entering and 9 mol exiting. We’ll need to multiply Ag3N4 by 3 to correct this imbalance. 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. By balancing everything else, we’ve also balanced out the nitrogen with 12 mol entering and 12 mol exiting. Next, we’ll need to correct our metals. Adjusting the NO31- amounts affected both the V and Ag amounts. Next, we’ll target the silver, Ag. There are 3 mol Ag entering the reaction and 1 mol exiting in Ag(NO3)4. Finally, we need to check out nitrogen, N. We’ll start out with V, vanadium. Technically, we could start out with Ag, silver, if you would like. Using the equation, only 1 mol vanadium enters as a reactant while 3 exits as a product. The LCM shared between 12 and 18 is 36. So, we’ll have to multiply the 3V(NO3)6 by 2 (3x2=6) and 3Ag(NO3)4 by 3 (3x3=9). So, 6V(NO3)6 and 9Ag(NO3)4. Now that we’ve balanced everything: 6 mol V, 36 mol NO31-, 9 mol Ag, and 12 mol N enter and exit this equation. Everything is properly balanced. To correct this, we’ll multiply the Ag(NO3)4 by a factor of 3. This will create the 3 mol of silver we need on both sides. 3 mol 9 mol 12 mol 3 mol 3 mol 1 mol Reactants: 18 mol 36 mol 3 mol 12 mol 1 mol 6 mol 3 mol Products: 12 mol 36 mol 9 mol 3 mol 6 mol Ag Ag V V N • Go back and double-check your work. Everything will balance if you are truly finished. • Work steadily through the reaction, element to element. • Typically, metals make good starting points for balancing chemical reactions. NO3(1-)

13. Practice On Your Own (I) • ___ RbOH + ___ H2SO4 ___ H2O + ___ Rb2SO4 • ___ C4H10 + ___ O2  ___ CO2 + ___ H2O • ___ CaCO3  ___ CaO + ___ CO2 • ___ CO2 + ___ H2O  ___ C6H12O6 + ___ O2 • ___ C8H18 + ___ O2  ___ CO2 + ___ H2O 2 1 2 1 2 8 13 10 1 1 1 6 6 1 6 2 25 16 18

14. Practice On Your Own (II) • ___ H2 + ____ O2 ___ H2O • ___ Pb(NO2)2 + ___ NaCl  ___ PbCl2 + ___ NaNO2 • ___ KClO3  ___ KCl + ___ O2 • ___ Fe + ___ HCl  ___ H2 + ___ FeCl3 • ___ C2H2 + ___ Cl2  ___ HCl + ___ C 2 1 2 1 2 1 2 2 2 3 2 2 6 3 1 2 1 2

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.