chemistry
posted by shadow55 on .
Please help =)
59. Ethanol, C2H5OH, is responsible for the effects of intoxication felt after drinking alcoholic beverages. When ethanol burns in oxygen, carbon dioxide and water produced.
(a) Write a balanced equation for the reaction.
[[Think I did this right]
Need help with these :
(b) How many liters of ethanol (d = 0.789 g/cm3) will produce 1.25 L of water (d = 1.00 g/cm3)?
(c) A wine cooler contains 4.5% ethanol by mass. Assuming that only the alcohol burns in oxygen, how many grams of wine cooler need to be burned to produce 3.12 L of CO2 (d = 1.80 g/L at 25°C, 1 atm pressure) at the conditions given the density?

It would have helped if you wrote the equation.
2C2H5OH + 5O2 ==> 4CO2 + 6H2O
1.25 L H2O has a mass of 1.00 gram.
1.00 g H2O = ?? mols H2O.
Convert mols H2O to mols C2H5OH using the coefficients in the balanced equation.
Convert mols ethanol to grams ethanol.
Using the density, convert g ethanol to liters ethanol.
Post your work if you get stuck.
I'll post the wine coolor problem separately. 
for the balanced equation i got 1C2H5OH + 3O2 ==> 2CO2 + 3H2O  why is this wrong?

I also don't completely understand this line:
1.25 L H2O has a mass of 1.00 gram.
I thought since the density was 1g /cm^3 it would be more like 1g / mL = 1000g/ L so 1250 g of H2O? 
Your equation is correct. I made a mistake when I did mine.

You are also correct about the water. 1.25 L x 1g/mL = 1,250 mL x 1.00 g/mL = 1,250 g H2O. I just misread the problem. It's past my bed time and I'm too sleepy. Good luck.

its okay its np =)

The wine cooler problem is the same as the ethanol except that the wine cooler is not 100% ethanol. It is only 4.5%. Therefore, first find the mols CO2 you want to produce. Use PV = nRT.
Convert mols CO2 to mols ethanol, then to grams ethanol. (I'll call the grams ethanol Y and use it below).
What you want is the number of grams of wine cooler (that is the ethanol PLUS the solvent) to produce Y grams ethanol.
(4.5*X/100) = Y where Y is the grams of ethanol from above and X is the grams of the wine cooler (the answer to your problem). Post your work if you get stuck. 
for (b) this is my work1250 g of H2O * 1 mol / 18 g = 69.44 mol of H2O
# of moles of C2H5OH = 69.44 mol of H2O * 1 mol of C2H5OH / 3 mol of H2O =
23.15 mol of C2H5OH * 46 g / 1 mol = 1065 g of C2H5OH
1065 g of C2H5OH * 1 cm3 / 0.789 g = 1350 cm3 of C2H5OH = 1350 mL of C2H5OH
= 1.35 L of C2H5OH
is this correct? 
I found 1350 cc also.