Category: Chemical Engineering Math
"Published in Newark, California, USA"
The vapor pressures of pure benzene and toluene at 60°C are 385 and 135 Torr, respectively. Calculate (a) the partial pressures of benzene and toluene, (b) the total vapor pressure of the solution, and (c) the mole fraction of toluene in the vapor above a solution with 0.60 mole fraction toluene.
Solution:
In the given problem, if mole fraction of toluene in a solution is
then mole fraction of benzene in a solution is
The sum of the mole fractions of the components in a solution must be equal to 1. In this case, benzene and toluene are the components in a solution.
The vapor pressures of pure benzene and toluene at 60°C are
(a) By Raoult's Law, we can calculate the partial pressures of benzene and toluene in the vapor.
Partial Pressure of Benzene:
Partial Pressure of Toluene:
(b) The total vapor pressure of the solution is
(c) Since we know already the partial pressures of benzene and toluene as well as the total vapor pressure of the solution, therefore, the mole fraction of toluene in the vapor is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
Four liters of octane gasoline weigh 3.19 kg. Calculate the volume of air required for its complete combustion at STP.
Solution:
The first thing that we have to do is to write the chemical equation for the combustion of octane gasoline as follows
Make sure that the above chemical equation is balanced. Next, we will calculate the molecular weight of octane gasoline as follows
Moles of octane gasoline:
From the chemical equation above, moles of oxygen is
At STP (Standard Temperature and Pressure), 1 mole of any gas is equal to 22.4 liters. Hence, the volume of oxygen is
Since air contains 21% oxygen and 79% nitrogen, therefore, the volume of air required for the complete combustion of octane gasoline is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
"Hard" water contains small amounts of the salt calcium bicarbonate [Ca(HCO3)2] and calcium sulfate [CaSO4, molecular weight = 136 grams/mole]. These react with soap before it has a chance to lather, which is responsible for its cleansing ability. Ca(HCO3)2 is removed by boiling to form insoluble CaCO3. CaSO4 is removed by reaction with washing soda [Na2CO3, molecular weight = 106 grams/mole] according to the following equation:
If the rivers surrounding New York City have a CaSO4 concentration of 1.8 x 10-3 grams/liter, how much Na2CO3 is required to "soften" [remove CaSO4] the water consumed by the city in one day [about 6.8 x 109 liters]?
Solution:
The first thing that we have to do is to get the amount of CaSO4 in the rivers surrounding New York City as follows
Moles of CaSO4:
From the given reaction above
Moles of Na2CO3:
Therefore, the amount of Na2CO3 required to "soften" or remove CaSO4 in the rivers surrounding New York City is
In metric tons, the weight of Na2CO3 is
