Going on from stoich we are now on limiting reactants. This concept is pretty basic. You start out with a given amount of grams for two elemnts and you have to figure out how much you can use to make a new one. Basiclly how much of each ingrediant you need to bake a cake. You then find one reactant which is limited. Lets go into a practice problem. Identify the limiting reactant in each of the following cases. 1.0 grams of h2 and 1.0 grams of O2. This is a two step solution.
First start with what you are given 1.0 grams of H2 multiply that by 1 mole over 2 grams of h2.(2 grams comes from finding the mass of H2). Next use the moles from the equation. Which would be 2 Mols of H20 over 2 moles of H2.(You dont get moles of 02 but H20 for both problems0. After doing the math you should get 0.5 mols of H20. Next is doing the second part except backwards. Start with 1.0 grams of 02 this time and to the same math. You would get .0625 mols of H2O. This would mean 02 is the limiting reactant because it is the lowest.
The problems get more complex like instead of moles you have to convert to grams so you just add and extra step add the end. whatever the grams of the elemt you are trying to find over 1 mol. Thats basiclly it. You are given two reactants and you have 2 problems. Just remeber to use the mole of the equation in the problems for the mass produced.
Homework-page 27 #2,page 29 #2 pages 30 and 31
Next scriber-Rachel Mitchel
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