Question

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{~kg} \mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction and assume that the NO produced in the third step is not recycled.

Short answer

The required mass of NH3 to produce \(1.0 \times 10^6\) kg of HNO3 using the Ostwald process is \(5.49 \times 10^8\) g or \(5.49 \times 10^5\) kg, assuming a 100% yield and no recycling of NO.

Step-by-step solution

Write the balanced chemical equations

The balanced chemical equations for the Ostwald process are given in the exercise: \[ \begin{aligned} 4 \mathrm{NH}_{3(g)} + 5 \mathrm{O}_{2(g)} \rightarrow 4 \mathrm{NO(g)} + 6 \mathrm{H}_{2}\mathrm{O(g)} \\ 2 \mathrm{NO(g)} + \mathrm{O}_{2(g)} \rightarrow 2 \mathrm{NO}_{2(g)} \\ 3 \mathrm{NO}_{2(g)} + \mathrm{H}_{2}\mathrm{O(l)} \rightarrow 2 \mathrm{HNO}_{3(aq)} + \mathrm{NO(g)} \end{aligned} \]

Find the moles of HNO3 produced

To find the moles of HNO3 produced, we'll use the equation: \(moles\ of\ HNO3 = \frac{mass\ of\ HNO3}{molar\ mass\ of\ HNO3}\) The molar mass of HNO3 (62.0 g/mol) can be determined from the periodic table by adding the molar masses of hydrogen, nitrogen, and oxygen, which are 1.0 g/mol, 14.0 g/mol, and 16.0 g/mol, respectively. The mass of HNO3 produced is given as 1.0 x 10^6 kg, which is equivalent to 1.0 x 10^9 g. Now we can find the moles of HNO3 produced: \(moles\ of\ HNO3 = \frac{1.0 \times 10^9\ g}{62.0\ g/mol} = 1.61 \times 10^7\ mol\)

Use stoichiometry to find the moles of NH3 required

To find the moles of NH3 required to produce the given amount of HNO3, we have to look at the stoichiometry of the overall reaction. For that, we need to add the three equations together: \[ 4 \mathrm{NH}_3 + 5 \mathrm{O}_2 + 2 \mathrm{NO} + \mathrm{O}_2 + 3 \mathrm{NO}_2 + \mathrm{H}_2\mathrm{O} \rightarrow 4 \mathrm{NO} + 6 \mathrm{H}_2\mathrm{O} + 2 \mathrm{NO}_2 + 2 \mathrm{HNO}_3 + \mathrm{NO} \] Canceling the common species on both sides of the equation, we get the net equation: \[ 4 \mathrm{NH}_3 + 9 \mathrm{H}_2\mathrm{O} \rightarrow 6 \mathrm{H}_2\mathrm{O} + 2 \mathrm{HNO}_3 \] Now, let's focus on the stoichiometric coefficients for NH3 and HNO3: \(4\ NH3 \rightarrow 2\ HNO3\) To find the moles of NH3 required to produce the given moles of HNO3, we can set up the following proportion: \(moles\ of\ NH3 = \frac{moles\ of\ HNO3 \times 4}{2}\) Now, we can plug in the moles of HNO3 we found in step 2: \(moles\ of\ NH3 = \frac{1.61 \times 10^7\ mol \times 4}{2} = 3.23 \times 10^7\ mol\)

Find the mass of NH3 required

To find the mass of NH3 required, we'll use the equation: \(mass\ of\ NH3 = moles\ of\ NH3 \times molar\ mass\ of\ NH3\) Using the periodic table, we can determine the molar mass of NH3 (17.0 g/mol) by adding the molar masses of nitrogen and hydrogen. Finally, we can find the mass of NH3 required: \(mass\ of\ NH3 = 3.23 \times 10^7\ mol\ NH3 \times 17.0 \frac{g}{mol\ NH3} = 5.49 \times 10^8\ g\) The required mass of NH3 to produce 1.0 x 10^6 kg HNO3 using the Ostwald process is \(5.49 \times 10^8\) g or \(5.49 \times 10^5\) kg, assuming a 100% yield and no recycling of NO.

Key concepts

Nitric Acid Production

The Ostwald process is the primary industrial method for producing nitric acid, a crucial component in fertilizers, explosives, and many other chemical products. This process starts with the catalytic oxidation of ammonia ( H_3 ext{)} to nitric oxide ( O ext{)} using a platinum-rhodium catalyst at high temperatures. Once formed, this nitric oxide is further oxidized to nitrogen dioxide ( O_2 ext{)}. Finally, the nitrogen dioxide reacts with water to produce nitric acid ( O_3 ext{)} and liberates nitric oxide, which can be reused in another cycle. Efficient recycling of NO means optimizing production, although in this exercise, it's supposedly not recycled. The entire sequence ensures high purity and efficiency in nitric acid production without generating unnecessary byproducts. Key factors influencing production efficiency include reaction temperature, pressure, and catalyst properties.

Stoichiometry

Stoichiometry is the heart of chemical reactions, allowing us to predict how much product we can obtain from given reactants. It involves examining the quantitative relationships between reactants and products in a chemical equation. In the context of this exercise, stoichiometry helps determine how much ammonia and oxygen are required to yield a specific amount of nitric acid. We use the stoichiometric coefficients in the balanced chemical equations to understand these relationships. For instance, the reactions within the Ostwald process reveal that 4 moles of ammonia are needed to ultimately produce 2 moles of nitric acid. By setting up stoichiometric calculations, we can efficiently calculate the reactant quantities necessary to produce the desired product. Grasping stoichiometry is pivotal for any aspiring chemist or engineer, as it guides efficient resource usage in industrial processes.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They provide crucial information about the reactants transforming into products and are balanced to maintain the conservation of mass, meaning that the number of atoms of each element is the same on both sides of the equation. In the Ostwald process, we have three chemical equations: 1. Ammonia reacting with oxygen to form nitric oxide and water. 2. Nitric oxide reacting with additional oxygen to form nitrogen dioxide. 3. Finally, nitrogen dioxide reacting with water to produce nitric acid and regenerate some nitric oxide. Each equation must be balanced. This ensures that the stoichiometry is correct, which is important for calculations involving quantities and masses of substances involved in the reactions. Understanding chemical equations enables us to predict the outcomes of reactions and the quantities of reactants needed for a desired amount of product.

Molar Mass Calculations

Molar mass calculations are fundamental to converting between grams and moles, a common requirement in chemical equations and stoichiometry. The molar mass is the mass of one mole of a substance in grams. It combines the atomic masses of the elements present in the compound. For instance, HNO_3 (nitric acid) has a molar mass calculated by combining the molar masses of hydrogen, nitrogen, and oxygen, resulting in 62 g/mol. Similarly, H_3 (ammonia) has a molar mass of 17 g/mol. To determine the mass of ammonia needed in the given exercise, its molar mass is multiplied by the calculated moles of H_3. Molar mass calculations ensure accurate measurements and conversions in chemistry, essential for both laboratory experiments and industrial applications, like the Ostwald process.