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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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Most popular questions from this chapter

At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A \(0.8765\)-\(\mathrm{g}\) sample of impure sodium chlorate was heated until the production of oxygen gas ceased. The oxygen gas collected over water occupied \(57.2 \mathrm{mL}\) at a temperature of \(22^{\circ} \mathrm{C}\) and a pressure of \(734\) torr. Calculate the mass percent of \(\mathrm{NaClO}_{3}\) in the original sample. (At \(22^{\circ} \mathrm{C}\) the vapor pressure of water is \(19.8\) torr.)

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\). This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus 1894 p. 1221 ) reacted beryllium with the anion \(C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows:If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .\) Show how Combes's data help to confirm that beryllium is a divalent metal.

Consider separate \(1.0-\mathrm{L}\) gaseous samples of \(\mathrm{H}_{2}, \mathrm{Xe}, \mathrm{Cl}_{2},\) and \(\mathbf{O}_{2}\) all at STP. a. Rank the gases in order of increasing average kinetic energy. b. Rank the gases in order of increasing average velocity. c. How can separate \(1.0-\mathrm{L}\) samples of \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2}\) each have the same average velocity?

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Complete the following table for an ideal gas. $$\begin{aligned} &-\\\ &\begin{array}{|llll|} \hline {P} & {V} & {n} & {\boldsymbol{T}} \\ \hline 7.74 \times 10^{3} \mathrm{Pa} & 12.2 \mathrm{mL} & & 25^{\circ} \mathrm{C} \\ \hline & 43.0 \mathrm{mL} & 0.421 \mathrm{mol} & 223 \mathrm{K} \\ \hline 455 \text { torr } & & 4.4 \times 10^{-2} \mathrm{mol} & 331^{\circ} \mathrm{C} \\ \hline 745 \mathrm{mm} \mathrm{Hg} & 11.2 \mathrm{L} & 0.401 \mathrm{mol} & \\ \hline \end{array} \end{aligned}$$
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