last check:
(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?
1 C2H5OH + 3 O2 = 2 CO2 + 3 H2O
3.12 L CO2 * (1.80 g / 1 L) = 5.62g CO2
5.62 g CO2 * (1 mol / 44 g CO2) = 0.128 mol of CO2
0.128 mol CO2 * (1 mol of C2H5OH / 2 mol of CO2) = 0.064 mol C2H5OH
0.064 mol of C2H5OH * (46 g / 1 mol) = 2.944g C2H5OH
2.944g C2H5OH * (100 / 4.5) =
65.42 g of wine cooler
Is this right? thanks
Is 1.80g/L the density of CO2 at standard conditions? If so, then that is correct, I believe.
I worked the problem and obtained 65.3. Your method is ok and the numbers look ok. I used molar masses that were a little different. Here is a site that does my calculation for me for molar masses.
http://environmentalchemistry.com/yogi/reference/molar.html
One final note. Your answer has four places in it but you are allowed only three since 3.12, 1.80 and 4.5(and I presume that is 4.50) have three places.
Great work guys! I read somewhere that people were using PV = nrt ... do you know why ?
Yes, your calculation is correct. To find the mass of wine cooler needed to produce 3.12 L of CO2, you followed the correct steps:
1. Convert the volume of CO2 to mass using the density: 3.12 L CO2 * (1.80 g / 1 L) = 5.62 g CO2
2. Convert the mass of CO2 to moles using the molar mass of CO2: 5.62 g CO2 * (1 mol / 44 g CO2) = 0.128 mol CO2
3. Use the stoichiometric ratio between C2H5OH (ethanol) and CO2 to find moles of C2H5OH: 0.128 mol CO2 * (1 mol of C2H5OH / 2 mol of CO2) = 0.064 mol C2H5OH
4. Convert moles of C2H5OH to mass using the molar mass of C2H5OH: 0.064 mol of C2H5OH * (46 g / 1 mol) = 2.944 g C2H5OH
5. Finally, convert the mass of C2H5OH to grams of wine cooler considering its concentration: 2.944 g C2H5OH * (100 / 4.5) ≈ 65.42 g of wine cooler.
Therefore, you would need approximately 65.42 grams of wine cooler to produce 3.12 L of CO2. Well done!