Chapter 8, Problem 8.116P

8-116 A railroad tank car derails and spills 26 tons of concentrated sulfuric acid (1 ton = 907.185 kg). The acid is 98.0% H₂SO₄ with a density of 1.836 g/mL.

(a) What is the molarity of the acid?

(b) Sodium carbonate, Na₂CO₃ is used to neutralize the acid spill. Determine the kilograms of sodium carbonate required to completely neutralize the acid. (Chapter 4)

(c) How many liters of carbon dioxide at 18°C and 745 mm Hg are produced by this reaction? (Chapter 5)

Interpretation Introduction

(a)

Interpretation:

The molarity of the sulfuric acid should be calculated.

Concept Introduction:

The molarity is the term used to describe the concentration of the aqueous solution. It is calculated by dividing the moles of a substance with volume of the substance in liters.

Answer to Problem 8.116P

The molarity of sulfuric acid is 18.36 M.

Explanation of Solution

The sulfuric acid is 98.0%. that means, if we have 100.0mL of 98.0% sulfuric acid with a density of 1.836g/mL, then the mass of the sulfuric acid would be

$\begin{array}{c}M=d\times V\\ =1.836\text{ g/mL}\times 100\text{ mL}\\ =183.6\text{ g}\end{array}$

Given that, 100.0 g of acid contains 98.0 g of sulfuric acid.

183.6 g acid contains.

$\begin{array}{c}\text{Total}\text{acid}=183.6\text{ g}\times 0.98\\ =179.93\text{ g}\end{array}$

The equation of molarity is

$\begin{array}{c}M=\frac{m}{M_m}\times \frac{1}{V(in\text{l})}\\ =\frac{179.93\mathrm{g}}{98\text{ g/mol}}\times \frac{1}{0.1\text{ L}}\\ =18.36\text{ M.}\end{array}$

Conclusion

The mass percentage and density of the acid are given. From we derived the mass of the acid present. Thus, molarity can be calculated and found to be 18.36 M.

Interpretation Introduction

(b)

Interpretation:

The amount of sodium carbonate required in kilograms to completely neutralize the acid should be calculated.

Concept Introduction:

The molar ratio of sulfuric acid and sodium carbonate is to be found. Then the amount of sodium carbonate required to neutralize the acid can be calculated easily.

Answer to Problem 8.116P

The mass of sodium carbonate required to completely neutralize the acid = 24,996.443 kg.

Explanation of Solution

The chemical reaction of sulfuric acid and sodium carbonate is represented using the following equation.

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\underset{}{\to }{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}+{\mathrm{CO}}_2$

From the chemical equation, it is clear that the molar ratio of sulfuric acid and sodium carbonate is 1:1.

Mass of sulfuric acid = 26 tons.

Mass of each ton = 907.185 kg.

Mass of sulfuric acid in kg = $26\times 907.185=23,586.81\text{ kg}$

But we know that the acid is 98.0% ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Thus, mass of sulfuric acid = $23,586.81\text{ kg}\times 0.98=23,115.0738\text{ kg}$

Moles of sulfuric acid are calculated as below.

$\begin{array}{c}n=\frac{M}{M}\\ =\frac{23,115.073\times {10}^3\text{ g}}{98\text{ g/mol}}\\ =235.86\times {10}^3\text{ mol}\end{array}$

From the chemical equation, the moles of sulfuric acid and sodium carbonate are in equal ratio. Hence, the number of moles of sodium carbonate is 235.86 $\times {10}^3$ mol. From the number of moles we can calculate the mass of sodium carbonate.

$\begin{array}{c}m=n\times M\\ =235.86\times {10}^3\text{mol}\times 105.98\text{ g/mol}\\ =24,996.443\times {10}^3\text{g}\\ =24,996.443\text{ kg}\end{array}$

Thus, the mass of sodium carbonate required to completely neutralize the acid = 24,996.443 kg.

Conclusion

The molar ratio of the reactants is used to calculate the mass of sodium carbonate. Thus, the mass of sodium carbonate required to completely neutralize the acid = 24,996.443 kg.

Interpretation Introduction

(c)

Interpretation:

The volume of carbon dioxide that is produced during the chemical reaction should be calculated.

Concept Introduction:

The molar ratio of sulfuric acid and carbon dioxide is to be found. The obtained amount of carbon dioxide is converted into volume using the ideal gas equation.

$PV=nRT$.

Answer to Problem 8.116P

The volume of carbon dioxide released during the chemical reaction = $5748.780\times {10}^3\text{L}$.

Explanation of Solution

The chemical reaction of sulfuric acid and sodium carbonate is represented using the following equation.

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\underset{}{\to }{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}$

From the chemical equation, it is clear that the molar ratio of sulfuric acid and carbon dioxide is 1:1.

Mass of sulfuric acid = 26 tons.

Mass of each ton = 907.185 kg.

Mass of sulfuric acid in kg = $26\times 907.185=23,586.81\text{ kg}$

But we know that the acid is 98.0% ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$. Thus,

Mass of sulfuric acid = $23,586.81\text{ kg}\times 0.98=23,115.0738\text{ kg}$

Moles of sulfuric acid are calculated as below.

$\begin{array}{c}n=\frac{m}{M}\\ =\frac{23,115.073\times {10}^3\text{ g}}{98\text{ g/mol}}\\ =235.86\times {10}^3\text{ mol}\end{array}$

From the chemical equation, the moles of sulfuric acid and carbon dioxide are in equal ratio. Hence, the number of moles of sodium carbonate is 235.86 $\times {10}^3$ mol. From the number of moles we can calculate the volume of carbon dioxide.

Pressure of the gas = 745 mm Hg.

We know that 760 mm Hg = 1 atm.

Thus, 745 mm Hg = 0.9802 atm.

Temperature of the gas = 18 $\text{C}$ =18 + 273 = 291 K.

The volume can be calculated using ideal gas equation.

$\begin{array}{c}PV=nRT\\ 0.9802\cancel{\text{atm}}\times V=235.86\times {10}^3\cancel{\text{mol}}\times 0.0821\text{L}\text{.}\cancel{\text{atm}}\text{/}\cancel{\text{mol}}.\cancel{\text{K}}\times 291\cancel{\mathrm{K}}\\ V=5748780.703\text{ L}\end{array}$

Thus, the volume of carbon dioxide released during the chemical reaction = $5748.780\times {10}^3\text{ L}$.

Conclusion

The molar ratio of the reactants is used to calculate the moles of carbon dioxide. Thus, the volume of carbon dioxide released during the chemical reaction = $5748.780\times {10}^3\text{ L}$.