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Chapter 5: Problem 164

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$\begin{array}{c}4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\\2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\\3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)\end{array}$$ 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

Expert verified
To produce 1.0 x 10鈦� kg of HNO鈧� by the Ostwald process, we first find the moles of HNO鈧� required as \(n_{\mathrm{HNO}_{3}}= (m_{\mathrm{HNO}_{3}} \times 1000) / M_{\mathrm{HNO}_{3}}\). Then, using stoichiometry and the balanced equations, we calculate the moles of NH鈧� required: \(n_{\mathrm{NH}_{3}} = (1/2) \times (3/2) \times n_{\mathrm{HNO}_{3}}\). Finally, we find the mass of NH鈧� required: \(m_{\mathrm{NH}_{3}} = n_{\mathrm{NH}_{3}} \times M_{\mathrm{NH}_{3}}\).

Step by step solution

01

Write down the given data and find the moles of HNO鈧�

We know that the mass of HNO鈧� (m鈧嶁倳鈧欋祾鈧冣値) we want to produce is 1.0 x 10鈦� kg. We can convert this to grams (1 kg = 1000 g) and then use the molar mass of HNO鈧� (M鈧嶁倳鈧欋祾鈧冣値 = 63 g/mol) to find the moles of HNO鈧� required. Moles of HNO鈧� (n鈧嶁倳鈧欋祾鈧冣値) = (m鈧嶁倳鈧欋祾鈧冣値 x 1000) / M鈧嶁倳鈧欋祾鈧冣値
02

Use stoichiometry with the balanced equations to find moles of NH鈧� required

Now, we're going to use stoichiometry with the balanced equations to determine the moles of NH鈧� required to produce the desired amount of HNO鈧�. From the third equation, we see that 3 moles of NO鈧� produce 2 moles of HNO鈧�. We can express this relationship using a ratio between moles of NO鈧� (n鈧嶁倷岬掆倐鈧�) and moles of HNO鈧� (n鈧嶁倳鈧欋祾鈧冣値): n鈧嶁倷岬掆倐鈧� = (3/2) 脳 n鈧嶁倳鈧欋祾鈧冣値 Next, from the second equation, we can find out how many moles of NO are required to produce the moles of NO鈧� calculated above: n鈧嶁倷岬掆倐鈧� = 2 脳 n_NO Now, let's find the moles of NH鈧� (n鈧嶁倷鈧曗們鈧�) required to produce the amount of moles of NO: From the first equation, we see that 4 moles of NH鈧� produce 4 moles of NO. We can express this relationship using a ratio between moles of NH鈧� (n鈧嶁倷鈧曗們鈧�) and moles of NO (n_NO) as follows: n鈧嶁倷鈧曗們鈧� = n_NO Putting everything together, we have: n鈧嶁倷鈧曗們鈧� = n_NO = (1/2) 脳 n鈧嶁倷岬掆倐鈧� = (1/2) 脳 (3/2) 脳 n鈧嶁倳鈧欋祾鈧冣値 Now, we can plug in the moles of HNO鈧� calculated in Step 1 to find the moles of NH鈧� required.
03

Convert the moles of NH鈧� to its mass

We have found the moles of NH鈧� required, now we can convert it to mass. Firstly, we will find the molar mass of NH鈧� (M鈧嶁倷鈧曗們鈧� = 17 g/mol). Mass of NH鈧� (m鈧嶁倷鈧曗們鈧�) = n鈧嶁倷鈧曗們鈧� 脳 M鈧嶁倷鈧曗們鈧� Using all this information, we can find the mass of NH鈧� required to produce 1.0 x 10鈦� kg of HNO鈧� by the Ostwald process.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ostwald process
The Ostwald process is an industrial method used for the production of nitric acid. A key component of this process involves the catalytic oxidation of ammonia. Several chemical reactions are used to accomplish this conversion. Each of these steps is carefully controlled to maximize the yield of nitric acid while minimizing energy consumption and waste. Here is a breakdown of the key reactions involved.
  • The first reaction oxidizes ammonia (NH extsubscript{3}) with oxygen (O extsubscript{2}) to produce nitrogen monoxide (NO) and water (H extsubscript{2}O).
  • The second reaction involves further oxidation of the nitrogen monoxide to nitrogen dioxide (NO extsubscript{2}).
  • Finally, nitrogen dioxide reacts with water to produce nitric acid (HNO extsubscript{3}) and some nitrogen monoxide is regenerated.
These reactions are exothermic, meaning they release energy, which can be utilized within the process to reduce external energy requirements. The Ostwald process allows for a continuous large-scale production of nitric acid, which is an essential chemical used across various industries.
Chemical equations
Chemical equations are a symbolic representation of a chemical reaction. They show reactants transforming into products, with coefficients that tell the ratio in which substances react or are produced. Understanding chemical equations is fundamental to solving problems involving stoichiometry, such as in the Ostwald process.In a balanced chemical equation:
  • The number of atoms for each element is the same on both sides of the equation.
  • Chemical symbols represent elements, while chemical formulas represent compounds.
For example, in the Ostwald process, each reaction in the sequence:\[4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g) \]is balanced, indicating the exact proportion of reactants needed and the products formed. By analyzing each equation step by step, we can calculate the amount of each reactant needed for a desired product yield, a concept deeply intertwined with stoichiometry.
Molar mass
Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It plays an essential role in converting mass into moles, which is a fundamental step in stoichiometry and chemical calculations such as those in the Ostwald process.To calculate the molar mass of a compound like nitric acid (HNO extsubscript{3}), simply add the atomic masses of the constituent atoms:
  • Hydrogen (H): 1.01 g/mol
  • Nitrogen (N): 14.01 g/mol
  • Oxygen (O): 16.00 g/mol
Thus, the molar mass of HNO extsubscript{3} is:\[M_{\mathrm{HNO}_{3}} = 1.01 + 14.01 + (3 \times 16.00) = 63.01 \text{ g/mol}\]In stoichiometric calculations, knowing the molar mass allows us to convert grams of a substance to moles, facilitating calculations of reactants and products required in a reaction.
Reaction yield
Reaction yield is an important concept in stoichiometry, as it describes the efficiency of a reaction. It can be expressed as a percentage to show the actual amount of product obtained from a given amount of reactant. A 100% reaction yield means all reactants have been converted to products without any loss.In industrial processes like the Ostwald process, maximizing yield is crucial to minimize waste and maximize efficiency. Reaction yield is calculated using the formula:\[\text{Reaction Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\]Understanding reaction yield allows chemists and engineers to optimize conditions such as temperature and pressure to enhance production. In scenarios where the yield is assumed to be 100%, as in the given exercise, the calculations are simplified, as no losses are accounted for. This assumption aids in theoretical calculations but must be adjusted for practical, real-world applications where yield could be less.

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Most popular questions from this chapter

Which of the following statements about chemical equations is(are) true? a. When balancing a chemical equation, you can never change the coefficient in front of any chemical formula. b. The coefficients in a balanced chemical equation refer to the number of grams of reactants and products. c. In a chemical equation, the reactants are on the right and the products are on the left. d. When balancing a chemical equation, you can never change the subscripts of any chemical formula. e. In chemical reactions, matter is neither created nor destroyed so a chemical equation must have the same number of atoms on both sides of the equation.

Commercial brass, an alloy of Zn and Cu, reacts with hydrochloric acid as follows: $$\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ (Cu does not react with HCl.) When 0.5065 g of a certain brass alloy is reacted with excess \(\mathrm{HCl}, 0.0985 \mathrm{g}\) \(\mathrm{ZnCl}_{2}\) is eventually isolated. a. What is the composition of the brass by mass? b. How could this result be checked without changing the above procedure?

Balance the following equations representing combustion reactions: c. \(C_{12} H_{22} O_{11}(s)+O_{2}(g) \rightarrow C O_{2}(g)+H_{2} O(g)\) d. \(\mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) e. \(\operatorname{FeO}(s)+\mathbf{O}_{2}(g) \rightarrow \operatorname{Fe}_{2} \mathbf{O}_{3}(s)\)

is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions. What mass of bacterial tissue is produced in a treatment plant for every \(1.0 \times 10^{4} \mathrm{kg}\) of wastewater containing \(3.0 \% \mathrm{NH}_{4}^{+}\) ions by mass? Assume that \(95 \%\) of the ammonium ions are consumed by the bacteria.

A sample of urea contains \(1.121 \mathrm{g} \mathrm{N}, 0.161 \mathrm{g} \mathrm{H}, 0.480 \mathrm{g} \mathrm{C}\) and \(0.640 \mathrm{g}\) O. What is the empirical formula of urea?
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