Can you find the mass of carbon dioxide released from a reaction?

Question

Use this equation:
C(subscript 2)H(sub 6)O +3O(sub 2) ------- 2CO(sub 2) +3H(sub 2)O

If 135g of ethanol are combusted with 101g O(sub 2), what is the mass (grams) of carbon dioxide released into the atmosphere? It is multiple choice, but I just want to know how to do the problem.
Thanks for your help!

Answer 1

First determine the limiting reagent. Calculate the moles of ethanol dividing the mass (135) by its molecular weight. Now, calculate the moles of O2 by dividing the mass of oxygen by its molecular weight (I think its 32) Compare these two results. Theoretically, the moles of ethanol must be one third that of the moles of oxygen. Then take the moles of the limiting reagent and watch the reaction stoichiometry, If ethanol is the limiting reagent, then the number of moles of CO2 will be as twice as much of ethanol. If the limiting reagent is oxygen, then the number of moles of CO2 will be two thirds of the oxygen. Then multiply the number of moles of CO2 by its molecular weight (44) in order to obtain the mass

Answer 2

first get the number of moles of each of the reactants to determine which one would be the limiting reactant: moles =Wt /MW

MW of C2H6O= (12)(2) +(1)(6)+(16)(1) = 46
MW of O2 = (16)(2) = 32

thus moles C2H6O = 135 / 46 = 2.93
moles of O2 = 101 / 32 = 3.1526

since there is less C2H6O then this is he limiting reactant. Thus, based form the chemical equation,

moles of CO2 = (2.93 moles C2H6O) (2moles CO2/1mole C2H6O) ... look at the constant before each element from the chem eq. The chem eq says that 1 mole of C2H6O plus 3 moles of O2 produces 2 moles of CO2 and 3moles of H20.

Now, moles of CO2 = 5.86.

Going back to the definition of moles: moles = wt/MW, to get the wt you multiply the moles with the MW, thus givnig you :

wt = moles x MW
= 5.86 x 44 (the MW of CO2)
= 257.84gm