How to find the limiting reagent
A simple step-by-step guide So let’s work with this example problem: Example: For the balanced equation shown below, what would be the limiting reagent if 67.3 grams of CaH2 were reacted with 74.9 grams of H2O? CaH2+2H2O=>Ca(OH)2+2H2 1 . 1. When finding the limiting reagent the only thing we are interested in is the first half of the equation given, so everything before the arrow is what we will concentrate on. Here we can first write down our 2 pieces of information: 67.3 grams of CaH2 74.9 grams of H2O 2. 2. Find the molar mass of each given piece of information Molar mass of CaH2 = 42 Molar mass of H2O = 17 If you do not know how to find the molar mass of a molecule, no fear… just click here. 3. 3. Divide the masses given in the equation by the molar masses you just found 67.3 grams of CaH2 / Molar mass of CaH2 -> 67.3/42 = 1.6 mol 74.9 grams of H2O / Molar mass of H2O -> 74.9/17 = 4.4 mol 4. 4. Now we need to look back to the original equation again so here it is: CaH2+2H2O=>Ca(OH)2+2H2 Look in front of the two molecules we have been focusing on and this is where you will find the ratio. - In this case CaH2 has no number in front of it so therefore it’s just 1 - 2H2O obviously has a 2 o so our ratio is 1:2 Finding the ratio is an extremely important part of finding the limiting reagent so if you are having trouble with this you can jus click here. 5. 5. Now we have the ratio we must compare and find the limiting reagent! CaH2 : 2H2O 1 : 2 To see which is the limiting just divide the number of moles by the number in front of the molecule: CaH2 1.6/1 = 1.6 mol 2H2O 4.4/2 = 2.2 mol 1.6 is obviously less than 2.2 so therefore…..CaH2 is the limiting reagent!
1.6 mol : 4.4 mol