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.
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?
9 answers
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?
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.
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?
# 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.
3 answers