posted by alex on .
for the reaction of 14.69 g of ethyl alcohol with 22.5 g of oxygen. which is the limiting reagent? how much CO2 will be produced? if 19.5 g of CO2 is actually produced, what is the percent yield? C2H5OH(1) + 3 O2 (g) - 2 CO2 (g) + 3H2O(g)
how much toluene (C7H8) is need to make 47.3 (g) of TNT (C7H5N3O6) by the equation given below? C7H8+ HNO3 - C7H5N3O6 + H2O
C2H5OH + 3O2 ==> 2CO2 + 3H2O
mols ethyl alc = grams/molar mass = about 0.319
mols O2 = 22.5/32 = about 0.7 but that's an estimate only.
Using the coefficients in the balanced equation, convert mols ethyl alcohol to mols CO2. That's approximately 0.64
Do the same for mols O2 to mols CO2. That's approximately 0.47. Of course these two answers are different which means one of them is wrong; the correct answer in limiting reagent problems is ALWAYS the smaller value and the limiting reagent is the one producint the smaller value of product. Then convert to grams CO2. g = mols x molar mass. This is the theoretical yield. Then
%yield = (actual yield/theor yield)*100
Your second problem is the same thing EXCEPT it isn't a limiting reagent problem. That means you just go through the above process one time and there is no comparing to see which is the limiting reagent. The toluene is the limiting reagent in the problem.