posted by starlight on .
(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.
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.