A chemist ran the reaction CH3CH2Cl(l) + KOH(aq) ^ CH3CH2OH(l) +KCL(aq) by heating 47.2 grams of CH3CH2Cl with excess KOH and obtained a yield of 28.7 grams of CH3CH2OH. What was his % yield?
I do not know where to begin with this problem. It only gives grams of one reactant? What is the actual yield? Detailed step by step instructions please.
The mass of the given reagent indicates that it is the limiting reagent, and the one that you base the theoretical yield on.
Find moles of CH3CH2Cl m/MM and then since it ...
This solution is provided in 141 words. It discusses how to find the percent yield using moles and the limiting reagent.