Anyone know anything about stoichiometry? It is kicking my backside. The problem is no matter how much the teacher goes over it I still find myself having problems in the set up of doing it. Setting it up to go from grams to atoms to moles and such. Once it's is set up the math is easy. It's just the set up that's giving me issues. Thanks Bob V
I think the 2 most important things to keep in mind while doing these types of problems are: 1) if you multiply every term in an equation by the same quantity, the equation is still true. For example, if you are given; 5x + 2y = 6z then it is also true that 10x + 4y = 12z (each term is multiplied by 2). 2) if you multiply any value by something that is equal to one, you do not change the overall value of that term. This is helpful because if two things are equal, setting them up as a fraction makes a fraction that is equal to one (anything divided by an equal value, as long as we are not talking about 0, is equal to 1). So if a=b, then a/b has to equal one as long as a and b do not equal 0. So how does any of this apply to stoichiometry problems? 1a) How many moles of chlorine gas (Cl2) would react with 5 moles of sodium (Na) according to the following chemical equation? (Balance equation.) Na + Cl2 --> NaCl First we have to balance the equation (sorry, but you're just going to have to imagine that the 2 after the Cl is in subscript). Because we are talking about chlorine gas (CL2) on the left side of the 'equation' and _Cl on the right side, we need a 2 in front of the NaCl because every 1 mole of Cl2 will yield 2 moles of NaCl. So at this point we have Na + Cl2 --> 2NaCl It's still not balanced yet though, because we are yielding 2 moles of NaCl from just 1 mole of Na. Easily taken care of by adding a 2 in front of the Na on the left 2Na + Cl2 --> 2NaCl To check and see if we are balanced, check how many moles of each element are represented on each side of the equation. We have 2 moles of sodium (Na) on the left and 2 moles of sodium on the right. Looks good there. Also 1 mole of chlorine gas (CL2) on the left which is a total of 2 moles of chlorine atoms, and 2 moles of chlorine on the right. So, the equation is balanced. We haven't yet answered the question (nor have we used either of the principles I laid out). How many moles of chlorine gas (Cl2) would react with 5 moles of sodium (Na)? In the balanced equation above, what do I need to multiply by in order to have a 5 in front of the sodium (Na) on the left? Answer - 2.5 (5/2). So, if I use prinicple #1 above, I need to multiply every term of the equation by 2.5: 5Na + 2.5Cl2 --> 5NaCl Which is read as "5 moles of Na plus 2.5 moles of Cl2 yields 5 moles of NaCl." The answer to the question is 2.5 moles of chlorine gas would react with 5 moles of sodium. 1b) Using the equation (after it is balanced) above, determine the amount of product that can be produced from 24.7 g Na. I'm guessing this is more the sort of problem that you were asking about that is giving you some trouble. First we need to know how many moles of Na is 24.7 grams of Na. This is where principle #2 comes in. We need to multiply 24.7 g Na by a fraction that is equal to one that will leave our term with units of moles of Na. Like this: 24.7 g Na x (___ moles Na/___ g Na) This way, the units of 'g Na' will cancel and we will be left with the units of 'moles of Na.' Using the periodic table, we know that 1 mole of Na is equal to 22.99 g Na. So, 24.7 g Na x (1 mole Na/22.9 g Na)= 1.0786 moles Na Now go back to the balanced equation 2Na + Cl2 --> 2NaCl What do I need to multiply each term by so that I get 1.0786 in front of the Na? Answer - 0.5393 (1.0786/2). When we multiply each term by 0.5393 (remember principle #1) we get 1.0786Na + 0.5393Cl2 --> 1.0786NaCl. So, determine the amount of product that can be produced from 24.7 g Na. We found that 24.7 g Na is equal to 1.0786 moles Na which will yield 1.0786 moles NaCl. Now we just need to figure out how many grams of NaCl that is. Time for principle #2 again. 1.0786 moles NaCl * (___ g NaCl/___ moles NaCl) = ____ g NaCl The units of moles NaCl will cancel leaving g NaCl. From the periodic table and the atomic weights of Na and Cl; 1 mole NaCl is equal to 58.44 g NaCl. 1.0786 moles NaCl * (58.44 g NaCl/1 mole NaCl) = 63 g NaCl 63 g NaCl can be produced from 24.7 g Na. 1c) How many molecules of product would be produced from 24.7g Na? We just found that 24.7g Na will produce 1.0786 moles of product (NaCl), so we use principle #2 above; 1.0786 moles NaCl * (____ molecules NaCl/____ moles NaCl) Here's where Avogadro comes in. By definition, 1 mole NaCl = 6.022x10^23 molecules of NaCl. 1.0786 moles NaCl * (6.022x10^23 molecules NaCl/1 mole Nacl) = 6.5x10^23 molecules NaCl. 6.5x10^23 molecules of product (NaCl) will be produced from 24.7g Na. Phew...long winded, but I hope it helps. If these are not the sort of stoichiometry problems that you were referring to, let me know and give a few examples and I'd be happy to try again.
My head hurts just reading it. I am glad that I don't have to worry about Stoichiometry after this semester. I found out from my Nuclear Medicine teacher we do not do any of that. It's only for the Intro to Chemistry class.
