Posted by Cyndy on Sunday, February 15, 2015 at 6:45pm.
How many grams of ammonium chloride are needed to raise the ammonia level 5 ppm in a 300 gallon system?
Now,
I have worked the problem so many times I have confused myself I think.
I keep getting 2 different answers depending upon how I work the question.
Idk if it's 5.6775 grams of NH4Cl or 16.82811 grams of NH4Cl since it takes 2.964 units of NH4CL to equal a 100% Ammonium solution
The way I read the problem it is a hydrolysis problem.
NH4^+ + H2O ==> NH3 + H3O^+
and you want the NH3 (ammonia, not ammonium) to be 5 ppm. Note that 5 ppm is 5E-3g/L and convert to mol/L and work the problem to see M NH4Cl. Then convert that to g NH4Cl in 300 gallons. The number you get for this seems excessively large but I think it is right; i.e., if you go at it another way and calculate the pH of that solution it gives you the right pH. That may be the point of the problem; i.e., to show you just how much it takes to raise the NH3 level as little as 5 ppm.
To calculate the number of grams of ammonium chloride needed to raise the ammonia level by 5 ppm in a 300-gallon system, you need to follow these steps:
Step 1: Convert the system's volume into liters.
Since you're given the volume in gallons, you need to convert it to liters. 1 gallon is approximately equal to 3.78541 liters. Hence, 300 gallons = 300 * 3.78541 liters = 1135.623 liters.
Step 2: Calculate the number of moles of ammonia needed to raise the level by 5 ppm.
To convert to moles, use the formula: amount (in moles) = concentration (in ppm) * volume (in liters) / 10^6.
In this case, the concentration is 5 ppm and the volume is 1135.623 liters.
So, moles of ammonia = (5 ppm * 1135.623 liters) / 10^6 = 5.678115 moles.
Step 3: Calculate the mass of ammonium chloride needed.
To find the mass of ammonium chloride, we need to consider the stoichiometry of the reaction. The balanced equation for this reaction is:
NH4Cl → NH3 + HCl.
From the equation, we can see that one mole of ammonium chloride is equivalent to one mole of ammonia. Hence, the number of moles of ammonium chloride needed is also 5.678115 moles.
Step 4: Convert moles to grams.
The molar mass of ammonium chloride (NH4Cl) is approximately 53.49146 g/mol.
Therefore, grams of ammonium chloride = 5.678115 moles * 53.49146 g/mol = 303.4292 grams.
Therefore, the correct answer is approximately 303.4292 grams of NH4Cl.
Note: In your calculation, you mentioned that it takes 2.964 units of NH4Cl to equal a 100% ammonium solution. However, since the question doesn't provide any information about the concentration of the ammonium solution or its purity, we assume that you need to directly add ammonium chloride to achieve the desired increase in ammonia level.