2.9 – Practice Problem 2

Question

Iron can be one of the most troublesome elements in water. It is very important to treat water containing large amounts of iron because as little as 0.3 ppm (parts per million) of iron can cause damage to manufacturing processes and the environment. The most common forms of iron found in water is ferrous, \(\text{Fe}^{+2}\), and ferric, \(\text{Fe}^{+3}\). The ferrous and ferric increases the acidity of the water. To treat the water, a base is needed such as sodium hydroxide, \(\text{NaOH}\). The base causes a titration with the ferrous titrating to iron hydroxide, \(\text{Fe(OH)}_2\), and ferric titrating to ferric hydroxide, \(\text{Fe(OH)}_3\). These hydroxides are not soluble in water and therefore can be extracted from the water.

\[\text{Fe}^{+2}_{(aq)} + 2 \space \text{NaOH}_{(aq)} \space \rightleftharpoons \space \text{Fe(OH)}_{2 \space (s)} + 2 \space \text{Na}_{(aq)}^{+1}\]

\[\text{Fe}^{+3}_{(aq)} + 3 \space \text{NaOH}_{(aq)} \space \rightleftharpoons \space \text{Fe(OH)}_{3 \space (s)} + 3 \space \text{Na}_{(aq)}^{+1}\]

You are a chemical engineer working at a mining company. You’re boss wants you to make a material balance of the water treatment of iron so that he can analyze the cost of the treatment. You will be using a continuous stirred-tank reactor (CSTR). Assume that only ferrous is present in the water.

\[\text{Fe}^{+2}_{(aq)} + 2 \space \text{NaOH}_{(aq)} \space \rightleftharpoons \space \text{Fe(OH)}_{2 \space (s)} + 2 \space \text{Na}_{(aq)}^{+1}\]

There are two input streams and two out put streams. The first input stream, stream 1, contains water, ferrous at 500 ppm. The flowrate of water in this stream is 450 L/h. The second input stream, stream 2, is a 3-molar solution of sodium hydroxide. The first output stream, stream 3, contains pure ferrous oxide. The second output stream, stream 4, contains all the water that entered the system, the sodium and the legal amount of iron in the water at 0.1 ppm.

1. Draw and label the body flow diagram (BFD).
2. Perform a degree of freedom analysis.

Answer

a) Draw and label the body flow diagram (BFD).

First, let’s draw the basic outline of the BFD

Attribution: Said Zaid-Alkailani & UBC [CC BY 4.0 de (https://creativecommons.org/licenses/by/4.0/)]

Now let’s turn everything into a molar basis and fill out our BFD.

\[\dot{V}_{H_2 O} = 450 \frac{L}{h}\]

\[\dot{n}_{H_2 O} = 450 \space \frac{L}{h} \times 1,000 \space \frac{g}{L} \times 0.0556 \space \frac{mol}{g} = 25,000 \space \frac{mol}{h}\]

Since we know that the ferrous in stream 1 is 500 ppm and the mole fractions of stream 1 must add to one, we find that:

\[y_{(1, Fe^{+2})} = 0.0005\]

\[y_{(1, H_2 O)} = 1 - y_{(1, Fe^{+2})} = 1 - 0.0005 = 0.9995\]

and that

\[\dot{n}_1 = \frac{\dot{n}_{H_2 O}}{y_{(1, H_2 O)}} = \frac{25,000 \space \frac{mol}{h}}{0.9995 } = 25.01 \space \frac{kmol}{h}\]

Since we know that only pure ferrous oxide exists stream there we find that:

\[y_{(1, Fe(OH)_2)} = 1\]

We can now label our BFD to the fullest extent without solving the problem.

Attribution: Said Zaid-Alkailani & UBC [CC BY 4.0 de (https://creativecommons.org/licenses/by/4.0/)]

b) Perform a degree of freedom analysis.

We have 8 unknown labeled variables but since we can deduce two of the mole fractions from the others, let’s just say we have 6 unknown labeled variables. We will use the atomic species balance for this question. The degree of freedom analysis for a atomic species balance is:

\[DOF = \text{Unknown variables} - \text{independent atomic species balance} - \text{molecular balances on independent nonreactive species} -\text{other equations relating unknown variables}\]

We know that:

\[\text{Unknown variables} = 6\]

\[\text{independent atomic species balance} = 4 \space (H, \space Fe, \space Na, \text{and} \space O)\]

\[\text{molecular balances on independent nonreactive species} = 0\]

\[\text{other equations relating unknown variables} = 2\]

Where other equations relating unknow variables is

\[y_{(4, Fe^{+2})} = 0.0001\]

and that \(NaOH\) is 3 moles per litre.

\[3 \space \frac{NaOH \space mol}{H_2 O \space l} \times \frac{ 1 \space l}{ 1,000 \space g} \times \frac{ 18 \space g}{1 \space mol \space H_2 O} = 0.054 \space \frac{NaOH \space mol}{H_2 O \space mol}\]

\[y_{(2, \space NaOH)} = \frac{0.054}{0.054 + 1} = 0.0512\]

\[\therefore \space DOF = 6 - 4 - 0 - 2 = 0\]