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.
To determine the limiting reagent, we need to compare the number of moles of each reactant (ethyl alcohol and oxygen) to determine which one will be completely consumed first.
1. Calculate the number of moles for each reactant:
- Moles of ethyl alcohol:
- Molar mass of C2H5OH (ethyl alcohol) = 46.07 g/mol
- Moles = mass / molar mass = 14.69 g / 46.07 g/mol
- Moles of oxygen:
- Molar mass of O2 (oxygen) = 32.00 g/mol
- Moles = mass / molar mass = 22.5 g / 32.00 g/mol
2. Determine the stoichiometric ratio:
- From the balanced equation, the ratio between ethyl alcohol and oxygen is 1:3. This means that for every 1 mole of ethyl alcohol, we need 3 moles of oxygen.
3. Compare the moles of reactants:
- Moles of ethyl alcohol = 14.69 g / 46.07 g/mol = 0.319 mol
- Moles of oxygen = 22.5 g / 32.00 g/mol = 0.703 mol
Since we have less moles of ethyl alcohol (0.319 mol) compared to oxygen (0.703 mol), the limiting reagent is ethyl alcohol. This means that ethyl alcohol will be completely consumed, and any remaining oxygen will be in excess.
4. Determine the amount of carbon dioxide produced:
- From the balanced equation, we know that it takes 2 moles of CO2 to produce 1 mole of ethyl alcohol:
- Moles of CO2 = 0.319 mol C2H5OH x (2 mol CO2 / 1 mol C2H5OH) = 0.638 mol CO2
5. Calculate the theoretical yield of CO2:
- The molar mass of CO2 = 44.01 g/mol
- Mass of CO2 = moles x molar mass = 0.638 mol x 44.01 g/mol = 28.06 g
So, the theoretical yield of CO2 is 28.06 g.
6. Calculate the percent yield:
- Percent yield = (actual yield / theoretical yield) x 100%
- Actual yield = 19.5 g (given in the question)
- Percent yield = (19.5 g / 28.06 g) x 100% = 69.48%
Therefore, the limiting reagent is ethyl alcohol, 28.06 g of CO2 will be produced theoretically, and the percent yield is 69.48%.