Chapter 4, Problem 138CP

a)

Interpretation Introduction

Interpretation: The total stream’s down flow rate has to be calculated.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

Answer to Problem 138CP

The total flow rate of downstream from the stream is $\text{5}{\text{.35×10}}^{\text{4}}\text{L/s}$ .

Explanation of Solution

Given:

Record the given data

Upstream at which the stream rate flows   = $\text{5}{\text{.00×10}}^{\text{4}}\text{L/s}$

Downstream at which the plant discharges= $\text{3}{\text{.50×10}}^{\text{3}}\text{L/s}$

Moles of $\text{HCl}$ in water                               = $\text{65}\text{.0}\text{ppm}$

The stream flow rate in upwards and downwards direction of a manufacturing plant and mole of $\text{HCl}$ in the downstream is recorded as shown above.

To calculate the stream’s total flow rate

$\text{Total}\text{flow}\text{rate=Upstream}\text{rate+downstream}\text{rate}$

                       = $\text{5}{\text{.00×10}}^{\text{4}}\text{L/s+3}{\text{.50×10}}^{\text{3}}\text{L/s}$

                       = $\text{5}{\text{.35×10}}^{\text{4}}\text{L/s}$

Total flow rate of the stream is $\text{5}{\text{.35×10}}^{\text{4}}\text{L/s}$

The total flow rate of the stream is calculated by summing up the values of upstream flow rate and the downstream flow rate. The total flow rate of the stream is $\text{5}{\text{.35×10}}^{\text{4}}\text{L/s}$.

b)

Interpretation Introduction

Interpretation: the concentration of $\text{HCl}$ has to be determined.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

Answer to Problem 138CP

Concentration of $\text{HCl}$ in ppm for downstream is $\text{4}\text{.25}\text{ppm}$

Explanation of Solution

Given:

Record the given data

Moles of $\text{HCl}$ in water = $\text{65}\text{.0}\text{ppm}$

Total flow rate of the plant  = $\text{5}{\text{.35×10}}^{\text{4}}\text{L/s}$

Downstream at which the plant discharges water= $\text{3}{\text{.50×10}}^{\text{3}}\text{L/s}$

The stream flow rate in upwards and downwards direction of a manufacturing plant and mole of $\text{HCl}$ in the downstream is recorded as shown above.

To calculate the concentration of $\text{HCl}$ in ppm downstream

$\begin{array}{l}\text{Concentration of HCl=}\frac{\text{3}{\text{.50×10}}^{\text{3}}\text{(65}\text{.0)}}{\text{5}{\text{.35×10}}^{\text{4}}}\\ \text{ =4}\text{.25}\text{ppm}\text{HCl}\end{array}$

Concentration of HCl downstream for the plant in ppm is 4.25 ppm

The concentration of HCl is calculated by plugging in the values of product of moles of HCl and downstream flow to the total flow rate of the plant. The concentration of HCl downstream for the plant in ppm is 4.25 ppm.

c)

Interpretation Introduction

Interpretation: The mass of $\text{CaO}$ consumed by the plant has to be determined.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

Answer to Problem 138CP

Mass of $\text{CaO}$ consumed in eight hours per day is $\text{1}{\text{.69×10}}^{\text{6}}\text{g}$

Explanation of Solution

Given:

Record the given data

Hours consumed                                                      = 8.00 hrs

Downstream stream flow rate for the second plant = 1.80×10⁴ L/s

Molar mass of HCl          = 36.46 g

Molar mass of CaO          =56.08 g

The hours consumed by CaO along with stream flow rate and molar masses of HCl and CaO are recorded as shown above.

To calculate the mass of CaO consumed by 8.00 hrs

The mass of CaO can be calculated from the mass of HCl

$\text{Mass}\text{of}\text{HCl=8}\text{.00}\text{h×}\frac{\text{60}\text{min}}{\text{h}}\text{×}\frac{\text{60}\text{s}}{\text{min}}\text{×}\frac{\text{1}{\text{.80×10}}^{\text{4}}\text{L}}{\text{s}}\text{×}\frac{\text{4}\text{.25}\text{mg}\text{HCl}}{\text{L}}\text{×}\frac{\text{1g}}{\text{1000}\text{mg}}\text{=2}{\text{.20×10}}^{\text{6}}\text{g}\text{HCl}$

Mass of HCl=2.20×10⁶ g

Therefore, the mass of CaO can be calculated by,

$\text{Mass}\text{of}\text{CaO=2}{\text{.20×10}}^{\text{6}}\text{g}\text{HCl×}\frac{\text{1}\text{mol}\text{HCl}}{\text{36}\text{.46}\text{g}\text{HCl}}\text{×}\frac{\text{1}\text{mol}\text{CaO}}{\text{2}\text{mol}\text{HCl}}\text{×}\frac{\text{56}\text{.08}\text{g}\text{Ca}}{\text{mol}\text{CaO}}\text{=1}{\text{.69×10}}^{\text{6}}\text{g}\text{CaO}$

Mass of CaO consumed in 8 hours work day by the plant is 1.69×10⁶ g

The mass of CaO consumed in 8 hrs work day by the plant is calculated by plugging in the values of mass of HCl with the molar masses of HCl and CaO to the flow rate of the downstream. The mass of CaO consumed is found to be 1.69×10⁶ g.

d)

Interpretation Introduction

Interpretation: the concentration of ${\text{Ca}}^{\text{2+}}$ in ppm of second plant downstream has to be determined.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

Answer to Problem 138CP

Concentration of ${\text{Ca}}^{\text{2+}}$ downstream for the second plant in ppm is $\text{10}\text{.3}\text{ppm}$

Explanation of Solution

Given:

Record the given data

Moles of calcium in the original stream = 10.2ppm

Mass of CaO           = 1.69×10⁶ g

Molar mass of CaO     = 56.08 g

Molar mass of Ca²⁺     = 40.08 g

Downstream stream flow rate for the second plant = 1.80×10⁴ L/s

Upstream at which the stream rate flows   = $\text{5}{\text{.00×10}}^{\text{4}}\text{L/s}$

Downstream at which the plant discharges= $\text{3}{\text{.50×10}}^{\text{3}}\text{L/s}$

Total flow rate of the plant  = $\text{5}{\text{.35×10}}^{\text{4}}\text{L/s}$

The molar masses of calcium and calcium oxide along with mass and moles of calcium oxide and calcium in stream along with total rate flow and the rate of upstream and downstream are recorded as shown above.

To calculate the concentration of Ca²⁺ in ppm downstream of the second plant if 90% of water is used.

$\begin{array}{l}\text{Concentration}\text{of}{\text{Ca}}^{\text{2+}}\text{going}\text{into}\text{the}\text{second}\text{plant=}\frac{\text{5}{\text{.00×10}}^{\text{4}}\text{(10}\text{.2)}}{\text{5}{\text{.35×10}}^{\text{4}}}\\ \\ \text{=9}\text{.53}\text{ppm}\\ \text{The second plant used= 1}{\text{.80×10}}^{\text{4}}\text{L/s×(8}\text{.00×60×60)s=5}{\text{.18×10}}^{\text{8}}\text{L}\text{of}\text{water}\\ \text{1}{\text{.69×10}}^{\text{6}}\text{g}\text{CaO×}\frac{\text{40}\text{.08}\text{g}{\text{Ca}}^{\text{2+}}}{\text{56}\text{.08}\text{g}\text{CaO}}\text{=1}{\text{.21×10}}^{\text{6}}\text{g}{\text{Ca}}^{\text{2+}}\text{was}\text{added}\text{to}\text{this}\text{water}\text{.}\\ \\ \text{Conecntration of}{\text{Ca}}^{\text{2+}}\text{=9}\text{.53+}\frac{\text{1}{\text{.21×10}}^{\text{9}}\text{mg}}{\text{5}{\text{.18×10}}^{\text{8}}\text{L}}\text{=9}\text{.53+2}\text{.34=11}\text{.87}\text{ppm}\end{array}$

$\begin{array}{l}\text{90}\text{.0\%}\text{of}\text{the}\text{water}\text{is}\text{returned,}\\ \text{Volume}\text{of}\text{water=(1}{\text{.80×10}}^{\text{4}}\text{)×0}\text{.900=1}{\text{.62×10}}^{\text{4}}\text{L/s}\text{of}\text{water}\\ \text{11}\text{.87 ppm}{\text{Ca}}^{\text{2+}}\text{is}\text{mixed}\text{with}\text{(5}\text{.35-1}{\text{.80)×10}}^{\text{4}}\text{=3}{\text{.55×10}}^{\text{4}}\text{L/s}\text{of}\text{water}\end{array}$

$\begin{array}{l}\text{3}{\text{.55×10}}^{\text{4}}\text{L/s}\text{contains}\text{9}\text{.53}\text{ppm}{\text{Ca}}^{\text{2+}}\\ \\ \text{Therefore,the}\text{final}\text{concentration}\text{of}{\text{Ca}}^{\text{2+}}\\ \\ \text{Final}\text{concentration}\text{of}{\text{Ca}}^{\text{2+}}\text{=}\frac{\text{(1}{\text{.62×10}}^{\text{4}}\text{L/s)(11}\text{.87}\text{ppm)+(3}{\text{.55×10}}^{\text{4}}\text{L/s)(9}\text{.53}\text{ppm)}}{\text{1}{\text{.62×10}}^{\text{4}}\text{L/s+3}{\text{.55×10}}^{\text{4}}\text{L/s}}\text{=10}\text{.3}\text{ppm}\end{array}$

The final concentration of Ca²⁺ returned by the second plant to the stream is 10.3 ppm

The concentration of Ca²⁺ if 90% of water is returned by the second plant to stream is calculated by using the concentration of Ca²⁺ before the water has been returned to the total volume. The final concentration of ${\text{Ca}}^{\text{2+}}$ is found to be $\text{10}\text{.3}\text{ppm}$