Free counters!

Tuesday, September 9, 2014

Stoichiometry Problem - Combustion

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



Monday, September 8, 2014

Stoichiometry Problem - Total Hardness

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



Sunday, September 7, 2014

Solubility Product Constant Problems

Category: Chemical Engineering Math

"Published in Newark, California, USA"

A chemist dissolves BaSO4 in pure water at 25°C. If Ksp = 1 x 10-10, what is the solubility of barium sulfate in water?

Solution:

Since barium sulfate is a slightly soluble salt in water, then it is not 100% completely dissolved. Usually, if a salt is slightly soluble in water, then the solution is cloudy at most. There's also a precipitate of undissolved salt at the bottom of the solution. Ksp will tell us how soluble the salt is. If Ksp = 0, then the salt is not soluble in water at all at any temperature. If you will shake the solution, it is cloudy at first and then later all particles will precipitate at the bottom of the solution. There are some salts that are more soluble in warmer solution rather than in room temperature solution. If Ksp = ∞, then the salt is 100% completely dissolved. The solution is clear at all at any temperature. 

Let's consider the barium sulfate that is slightly soluble in water


The solubility of barium sulfate is calculated by its equation


The concentration of barium ion and sulfate ion are expressed in molarity.

Let x = be the concentration of barium ion in the solution.
      x = be the concentration of sulfate ion also in the solution.

Therefore, the solubility of barium sulfate in water is