I made sure to do lots of practice questions in the multiple packets I printed off for this unit. In summary for Stoichiometry you basically find the limiting reagent and excess reagent. To do that you must take 2 elements/compounds with their given mass in grams. You must convert it both to moles in separate equations and then multiply that by the number of moles (coefficients!!!) for the element/compound you are trying to find and the number of moles of the element you are working on on the bottom of the fraction. Then you multiply that by the molar mass of the element/compound you are trying to find and put that over 1 mole. Once that is done, you look at which element/compound that produced the least amount of whatever element/compound that you're trying to find. The least one is your limiting reagent because that element/compound will run out first and limit the amount you can make in the other element/compound that you found. That will be your answer! :)
In the excess one (the element that produced the most amount of the element/compound), you basically do the same thing. However you take the limiting reagent amount (smallest number you found above) and convert that to moles. Then you multiply it by the molar fraction with the moles of the excess reagent on top and the moles of the element/compound that you were solving for on the bottom. Then you multiply that by the molar mass of the excess reagent. Finally you take the starting mass of the excess reagent subtracted by the number you just found. VOILA! You are now an expert! :)
Helpful links below!
STOICHIOMETRY - Limiting Reactant & Excess Reactant Video
Introduction to Limiting Reactant and Excess Reactant
Finding Limiting Reagent
No comments:
Post a Comment