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?

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?

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 =)

I found 1350 cc also.

(a) The balanced equation for the combustion of ethanol is:

C2H5OH + 3O2 -> 2CO2 + 3H2O

(b) To solve this problem, we need to use the density and volume of water to determine the mass of water produced, and then use stoichiometry to find the amount of ethanol required to produce that much water.

First, let's calculate the mass of water produced:
Given: Density of water (d) = 1.00 g/cm^3
Volume of water (V) = 1.25 L

Mass of water = Volume * Density = 1.25 L * 1.00 g/cm^3 = 1.25 kg

Now, using stoichiometry, we can find the amount of ethanol required to produce 1.25 kg of water:
From the balanced equation, we know that 1 mole of ethanol produces 3 moles of water.

Molar mass of ethanol (C2H5OH) = (12.01 g/mol * 2) + (1.01 g/mol * 6) + (16.00 g/mol) = 46.07 g/mol

Number of moles of water produced = (1.25 kg) / (18.02 g/mol) = 69.38 mol

Number of moles of ethanol required = (69.38 mol water) / (3 mol water/mol ethanol) = 23.13 mol

To convert moles to grams, multiply by the molar mass:
Mass of ethanol required = 23.13 mol * 46.07 g/mol = 1065.82 g

Finally, we need to convert grams to liters using the density of ethanol:
Given: Density of ethanol (d) = 0.789 g/cm^3

Volume of ethanol = Mass / Density = 1065.82 g / 0.789 g/cm^3 = 1351.92 cm^3

Since 1 cm^3 is equal to 1 mL, the volume of ethanol is also 1351.92 mL.

Therefore, 1351.92 mL of ethanol will produce 1.25 L of water.

(c) To solve this problem, we need to find the molar mass of ethanol and the mass of the wine cooler, and then use stoichiometry to find the mass of the wine cooler needed to produce a given volume of CO2.

First, let's calculate the molar mass of ethanol:
Molar mass of ethanol (C2H5OH) = (12.01 g/mol * 2) + (1.01 g/mol * 6) + (16.00 g/mol) = 46.07 g/mol

Next, let's calculate the mass of ethanol in the wine cooler:
Given: Mass percent of ethanol = 4.5%
Let's assume we have 100 grams of wine cooler.

Mass of ethanol in the wine cooler = (4.5 / 100) * 100 grams = 4.5 grams

Now, using stoichiometry, we can find the mass of wine cooler needed to produce a given volume of CO2.

Given: Volume of CO2 (V) = 3.12 L
Density of CO2 (d) = 1.80 g/L

Mass of CO2 = Volume * Density = 3.12 L * 1.80 g/L = 5.62 g

From the balanced equation, we know that for every 2 moles of ethanol burned, 2 moles of CO2 are produced.

Number of moles of CO2 produced = (5.62 g CO2) / (44.01 g/mol) = 0.1276 mol CO2

Number of moles of ethanol required = 0.1276 mol CO2

Mass of wine cooler needed = (0.1276 mol CO2) * (46.07 g/mol ethanol) = 5.89 g

Therefore, 5.89 grams of wine cooler need to be burned to produce 3.12 L of CO2.