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Chapter 8: Problem 75

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g),\) ammonia, \(\mathrm{NH}_{3}(g),\) and oxygen, \(\mathrm{O}_{2}(g),\) at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of HCN( \(g\) ) can be obtained from the reaction of \(20.0 \mathrm{L} \mathrm{CH}_{4}(g), 20.0 \mathrm{L} \mathrm{NH}_{3}(g),\) and \(20.0 \mathrm{LO}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.

Short Answer

Expert verified
The balanced chemical equation for the reaction is \(CH_4(g) + NH_3(g) + \frac{3}{2}O_2(g) \rightarrow HCN(g) + 3H_2O(g)\). When 20.0 L of each reactant is provided at the same temperature and pressure, oxygen is the limiting reagent, and the volume of HCN that can be obtained from the reaction is 13.33 L.

Step by step solution

01

Write the chemical equation

For the reaction of methane, ammonia, and oxygen, with hydrogen cyanide and gaseous water as the products, we write: \(CH_4(g) + NH_3(g) + O_2(g) \rightarrow HCN(g) + H_2O(g)\)
02

Balance the chemical equation

To balance the equation, we make sure that there are the same number of atoms on both sides of the equation: \(CH_4(g) + NH_3(g) +3/2\, O_2(g) \rightarrow HCN(g) + 3\, H_2O(g)\) Now, the balanced chemical equation is: \(CH_4(g) + NH_3(g) + \frac{3}{2}O_2(g) \rightarrow HCN(g) + 3H_2O(g)\) #b. Determine the volume of HCN produced from the reaction#
03

Determine the limiting reagent

Since all gases have the same temperature and pressure, we can use volume ratios directly. From the balanced chemical equation, the volume ratios are: \(CH_4 : NH_3 : O_2 : HCN : H_2O = 1 : 1 : \frac{3}{2} : 1 : 3\) Given that we have \(20.0 L\) of each reagent, we determine which reagent will be consumed first, making it the limiting reagent. For methane, \(20.0 L \, CH_4 \times \frac{1}{1} = 20.0 L\) For ammonia, \(20.0 L \, NH_3 \times \frac{1}{1} = 20.0 L\) For oxygen, \(20.0 L \, O_2 \times \frac{2}{3} = 13.33 L\) Since 13.33 L of oxygen is consumed, oxygen is the limiting reagent.
04

Calculate the volume of HCN produced by the limiting reagent

Now we use the volume ratio to calculate the volume of HCN produced. From the balanced chemical equation, the ratio of volumes is: \(O_2 : HCN = \frac{3}{2} : 1\) Now we can calculate the volume of HCN produced: \(Volume \,of\, HCN = 20.0 L \, O_2 \times \frac{1}{\frac{3}{2}} = 13.33 L\) So, the volume of HCN that can be obtained from the reaction is \(13.33 L\).

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Key Concepts

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

Stoichiometry
Stoichiometry is like the cookbook of chemistry. It tells us the exact recipe of reactants to mix in order to get the desired chemical products. Imagine you want to bake cookies, and the recipe says for every cup of sugar, you need two cups of flour. That's stoichiometry! In chemical reactions, stoichiometry is used to calculate the amounts of reactants and products.
  • Stoichiometry deals with the quantitative relationships between the reactants and products in a chemical reaction.
  • It uses molar relationships and balanced chemical equations to find out how much of one chemical is needed to react with another.

To solve problems using stoichiometry, start by balancing the chemical equation. Then, use the molar ratios from the balanced equation to convert between different substances in the reaction. The goal is to find exactly how much product you can make from a given amount of reactant, as shown in the original exercise.
Limiting Reagent
The limiting reagent is the true boss of any chemical reaction. It decides when the reaction stops. It's like having an equal number of chairs and guests at a dinner party. Once all chairs are filled, no more guests can sit. The guest count here is controlled by the limiting reagent.
  • This reagent is completely consumed in a reaction and prevents more products from being created.
  • It determines the maximum amount of product a chemical reaction can produce.

In the exercise, oxygen was the limiting reagent because it was the first one to be completely used up. While there was enough methane and ammonia, the available oxygen limited the creation of additional hydrogen cyanide and water. This is why calculating which reactant is limiting is crucial to figuring out how much product you can actually obtain. Once you've identified the limiting reagent, you use it to calculate the amount of product that can be formed.
Balanced Chemical Equation
A balanced chemical equation is the cornerstone of every stoichiometry problem. Imagine trying to assemble a Lego set without the instructions鈥攊t would be chaotic. Balancing chemical equations ensures that the same number of each type of atom is present on both sides of the reaction equation. This mirrors the Law of Conservation of Mass, which states that matter cannot be created or destroyed.
  • Products and reactants in a balanced equation follow fixed proportions, ensuring mass is conserved during the reaction.
  • The coefficients in a balanced equation are essential鈥攖hey tell us the ratio in which chemicals react.

In the exercise, we balanced the equation to determine the ratios of methane, ammonia, and oxygen needed to produce hydrogen cyanide and water. Without balancing the equation, predicting the outcome and quantities of products would be impossible. Always balance your chemical equations before proceeding with stoichiometric calculations to ensure accuracy.

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Most popular questions from this chapter

Calculate the average kinetic energies of \(\mathrm{CH}_{4}(g)\) and \(\mathrm{N}_{2}(g)\) molecules at \(273 \mathrm{K}\) and \(546 \mathrm{K}\).

Silane, \(\mathrm{SiH}_{4},\) is the silicon analogue of methane, \(\mathrm{CH}_{4} .\) It is prepared industrially according to the following equations: $$\begin{aligned} \mathrm{Si}(s)+3 \mathrm{HCl}(g) & \longrightarrow \mathrm{HSiCl}(i)+\mathrm{H}_{2}(g) \\\4 \mathrm{HSiCl}_{3}(l) & \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l)\end{aligned}$$ a. If \(156 \mathrm{mL} \mathrm{HSiCl}_{3}(d=1.34 \mathrm{g} / \mathrm{mL})\) is isolated when \(15.0 \mathrm{L}\) HCl at \(10.0\) atm and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of HSiCl_? b. When \(156 \mathrm{mL}\) \(HSiCl_{3}\)is heated, what volume of \(\mathrm{SiH}_{4}\) at \(10.0\) atm and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is \(93.1 \% ?\)

Urea \(\left(\mathrm{H}_{2} \mathrm{NCONH}_{2}\right)\) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide: $$2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{NCONH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)$$ Ammonia gas at \(223^{\circ} \mathrm{C}\) and \(90 .\) atm flows into a reactor at a rate of \(500 .\) L/min. Carbon dioxide at \(223^{\circ} \mathrm{C}\) and \(45\) atm flows into the reactor at a rate of \(600 .\) L/min. What mass of urea is produced per minute by this reaction assuming \(100 \%\) yield?

Equal moles of sulfur dioxide gas and oxygen gas are mixed in a flexible reaction vessel and then sparked to initiate the formation of gaseous sulfur trioxide. Assuming that the reaction goes to completion, what is the ratio of the final volume of the gas mixture to the initial volume of the gas mixture if both volumes are measured at the same temperature and pressure?

Which of the following statements is(are) true? a. If the number of moles of a gas is doubled, the volume will double, assuming the pressure and temperature of the gas remain constant. b. If the temperature of a gas increases from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) the volume of the gas would double, assuming that the pressure and the number of moles of gas remain constant. c. The device that measures atmospheric pressure is called a barometer. d. If the volume of a gas decreases by one half, then the pressure would double, assuming that the number of moles and the temperature of the gas remain constant.
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