Answer to Question #61201 in General Chemistry for mae

Question

How many milliliters of air are needed to oxidize 4 lbs of butyl butyrate (C8H16O)? Find the number of grams of nitrogen gas present in the air for the combustion. What is the volume reacted (in liters) for nitrogen gas to oxidize butyl butyrate? If the air that will be used for the oxidation of butyl butyrate is a wet air and has 180 grams of moisture (H2O vapor), find the volume of the wet air (in liters) at STP.

Answer 1

First of all, the formula of butyl butyrate is incorrect. The correct formula contains 2 oxygens - C8H16O2.
Now, we are calculating the mass of butyl butyrate in kg - 4 lbs*0.454 kg/lbs = 1.816 kg.
The combustion can be written as: C8H16O2 + 11O2 = 8CO2 + 8H2O
According to this equation 1.816 kg or 0.0126 mol of butyl butyrate requires 0.1386 mol or 3.105 liters of pure oxygen.
The volume content of oxygen in the atmosphere is 20.946 %. Thus, the total volume of air is 14.824 liters (14824 ml, STP).
The volume reacted of nitrogen is 11.575 liters. Its molar quantity is 0.517 mol or 14.5 g.
If the air is wet and the total volume of water vapor is 224 liters (STP), the total volume of the wet air is a sum of the dry air and the moisture volume: 224 liters + 15 liters = 239 liters.