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Chapter 4: Problem 53

Ammonia is oxidized to nitric oxide in the following reaction: $$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$$ (a) Calculate the ratio (lb-mole \(\mathrm{O}_{2}\) react/lb-mole NO formed). (b) If ammonia is fed to a continuous reactor at a rate of \(100.0 \mathrm{kmol} \mathrm{NH}_{3} / \mathrm{h}\), what oxygen feed rate (kmol/h) would correspond to 40.0\% excess O_? (c) If \(50.0 \mathrm{kg}\) of ammonia and \(100.0 \mathrm{kg}\) of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced (kg) if the reaction proceeds to completion.

Short Answer

Expert verified
(a) The ratio is 1.25 lb-mole \(\mathrm{O}_{2}\) react per lb-mole NO formed. (b) The \(\mathrm{O}_{2}\) feed rate would be 175 kmol/h. (c) The limiting reactant is \(\mathrm{O}_{2}\), \(\mathrm{NH}_{3}\) is in excess by 17.64%, and the extent of reaction allows for the production of 75 kg of \(\mathrm{NO}\).

Step by step solution

01

Calculate the molar ratio

From the balanced equation, the ratio \(\mathrm{O}_{2}\) to \(\mathrm{NO}\) is \(\frac{5 \mathrm{O}_{2}}{4\mathrm{NO}}\). This simplifies to a ratio of \(\frac{5}{4}\) or 1.25.
02

Calculating the oxygen feed rate

The \(\mathrm{NH}_{3}\)'s feed rate is 100.0 kmol per hour with an excess of 40% \(\mathrm{O}_{2}\). The stoichiometric feed rate of \(\mathrm{O}_{2}\) is 1.25 times the feed rate of \(\mathrm{NH}_{3}\) (from Step 1) or \(1.25 \times100 = 125\) kmol/h. With a 40\% excess, the \(\mathrm{O}_{2}\)'s feed rate is \(125 + (0.40 \times 125) = 175\) kmol/h.
03

Determining the limiting reactant

If 50.0 kg of \(\mathrm{NH}_{3}\) and 100.0 kg of \(\mathrm{O}_{2}\) are fed into the reactor, we need to convert the masses to moles. Using the molar masses (\(\mathrm{NH}_{3}\)'s is approximately 17 g/mol and \(\mathrm{O}_{2}\)'s is approximately 32 g/mol), we get \(2941\) moles of \(\mathrm{NH}_{3}\) and \(3125\) moles of \(\mathrm{O}_{2}\). The molar ratio of \(\mathrm{O}_{2}:NH_{3}\) from the reaction is \(5:4\). So for every 4 moles of \(\mathrm{NH}_{3}\), we need 5 moles of \(\mathrm{O}_{2}\). So the amount of \(\mathrm{O}_{2}\) needed is \(5/4 \times 2941=3676\) moles. As we have only \(3125\) moles of \(\mathrm{O}_{2}\) present, \(\mathrm{O}_{2}\) is the limiting reactant.
04

Calculating the Excess Reactant

Having figured out that \(\mathrm{O}_{2}\) is the limiting reactant, we need to discern how much \(\mathrm{NH}_{3}\) is in excess. From the equation, we know that for every 5 moles of \(\mathrm{O}_{2}\), 4 moles of \(\mathrm{NH}_{3}\) are required. So the number of moles of \(\mathrm{NH}_{3}\) needed is \(4/5 \times 3125=2500\) moles. Hence, \(2941-2500 = 441\) moles of \(\mathrm{NH}_{3}\) are in excess. The percentage by which \(\mathrm{NH}_{3}\) is in excess is then \((441/2500) \times 100\%\), or \(17.64\%.\)
05

Calculating the Extent of reaction and Mass of NO produced

The limiting reactant dictates the reaction's extent. Given that \(\mathrm{O}_{2}\) is the limiting reactant, the reaction can't go beyond \(3125\) moles. To calculate the mass of \(\mathrm{NO}\) produced, we can use the balanced equation, which tells us that 4 moles of \(\mathrm{NH}_{3}\) yield 4 moles of \(\mathrm{NO}\) - which yields a 1:1 ratio. Therefore, \(2500\) moles of \(\mathrm{NH}_{3}\) (the amount that reacted) yield \(2500\) moles of \(\mathrm{NO}\). To convert this to mass, multiply by \(\mathrm{NO}\)'s molar mass (\(30\) g/mol), yielding \(2500 * 30 = 75000\) g or \(75\) kg of \(\mathrm{NO}\).

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Key Concepts

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

Limiting Reactant
Understanding the concept of a limiting reactant is crucial in chemical reactions, as it determines how much product can be formed. A limiting reactant is the reactant that is completely consumed first in a chemical reaction, thus limiting the amount of product that can be produced. It's analogous to running out of gas on a road trip; no matter how much of the other reactants (or snacks) you have left, the trip can't continue without gas.

Determining the limiting reactant involves comparing the molar amount of all reactants based on the reaction's stoichiometry. In the given problem where ammonia (\(NH_{3}\)) is oxidized to nitric oxide (\(NO\) ), if we start with 50.0 kg of \(NH_{3}\) and 100.0 kg of \(O_{2}\), we first convert the mass of the reactants to moles. Once we have the moles, we can use the balanced chemical equation to find out the ideal ratio between \(NH_{3}\) and \(O_{2}\) to react completely without any remainder. In this case, \(O_{2}\) runs out before \(NH_{3}\), making \(O_{2}\) the limiting reactant.

When teaching this concept, it can be helpful to visually illustrate the process using diagrams that show mole comparisons, ensuring students understand that it's not necessarily about the 'amount' of a substance, but about the ratio in which it reacts based on the balanced chemical equation.
Excess Reactant
In contrast to the limiting reactant, we have an excess reactant, which is the substance that remains after the reaction has gone to completion because there was more of it than necessary according to the stoichiometric ratios. Once the limiting reactant is entirely used up, the reaction stops, and it's not possible to use the remaining excess reactant without adding more of the limiting one.

In our exercise, after determining that \(O_{2}\) is the limiting reactant, we calculate the amount of the excess reactant, \(NH_{3}\), which was not consumed during the reaction. This is important for two reasons: it can be recovered and used elsewhere, and it informs adjustments for future reactions to minimize waste and costs. By calculating the exact excess percentage of \(NH_{3}\), we provide valuable information for optimizing the reaction's efficiency in practical applications.

Students might benefit from practical examples or experiments that demonstrate how excess reactants can be quantified in a lab setting and discussing why this could be economically and environmentally significant.
Molar Ratio
The molar ratio is a foundational concept in stoichiometry that expresses the proportions of reactants and products in a balanced chemical reaction. It is derived from the coefficients of each substance in the balanced equation and informs us exactly how much of one reactant is needed to react with another. This is the stoichiometric 'recipe' for the reaction to occur perfectly without any leftover reactants.

For the exercise presented, the molar ratio of \(O_{2}\) to \(NO\) from the balanced equation is 1.25, indicating 1.25 moles of oxygen are required for every mole of nitric oxide produced. When working with a continuous reactor scenario, as in question (b), understanding molar ratios allows us to predict the amount of one reactant (oxygen) needed when we're given the flow rate of another (ammonia), including accounting for excess to ensure complete reaction.

When teaching molar ratios, it is effective to engage students in mole-to-mole conversion practice and help them visualize the concept with pie charts or other graphical representations that show the proportion of reactants to products. This aids them in developing an instinctive understanding of the quantities involved in reactions.

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Most popular questions from this chapter

The gas-phase reaction between methanol and acetic acid to form methyl acetate and water takes place in a batch reactor. When the reaction mixture comes to equilibrium, the mole fractions of the four reactive species are related by the reaction equilibrium constant $$K_{y}=\frac{y_{C} y_{D}}{y_{A} y_{B}}=4.87$$ (a) Suppose the feed to the reactor consists of \(n_{\mathrm{A} 0}, n_{\mathrm{B} 0}, n_{\mathrm{C} 0}, n_{\mathrm{D} 0},\) and \(n_{10}\) gram-moles of \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and an inert gas, I, respectively. Let \(\xi\) be the extent of reaction. Write expressions for the gram-moles of each reactive species in the final product, \(n_{\mathrm{A}}(\xi), n_{\mathrm{B}}(\xi), n_{\mathrm{C}}(\xi),\) and \(n_{\mathrm{D}}(\xi) .\) Then use these expressions and the given equilibrium constant to derive an equation for \(\xi_{c}\), the equilibrium extent of reaction, in terms of \(\left.n_{\mathrm{A} 0}, \ldots, n_{10} . \text { (see Example } 4.6-2 .\right)\) (b) If the feed to the reactor contains equimolar quantities of methanol and acetic acid and no other species, calculate the equilibrium fractional conversion. (c) It is desired to produce 70 mol of methyl acetate starting with 75 mol of methanol. If the reaction proceeds to equilibrium, how much acetic acid must be fed? What is the composition of the final product? (d) Suppose it is important to reduce the concentration of methanol by making its conversion at equilibrium as high as possible, say 99\%. Again assuming the feed to the reactor contains only methanol and acetic acid and that it is desired to produce 70 mol of methyl acetate, determine the extent of reaction and quantities of methanol and acetic acid that must be fed to the reactor. (e) If you wanted to carry out the process of Part (b) or (c) commercially, what would you need to know besides the equilibrium composition to determine whether the process would be profitable? (List several things.)

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