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Chapter 3: Problem 168

A gas contains a mixture of \(\mathrm{NH}_{3}(g)\) and \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})\), both of which react with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{NO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\). The gaseous mixture (with an initial mass of \(61.00 \mathrm{~g}\) ) is reacted with \(10.00\) moles \(\mathrm{O}_{2}\), and after the reaction is complete, \(4.062\) moles of \(\mathrm{O}_{2}\) remains. Calculate the mass percent of \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})\) in the original gaseous mixture.

Short Answer

Expert verified
The mass percent of N₂H₄ in the original gaseous mixture cannot be determined with the given data, as the calculated mass percent is greater than 100%. This indicates an error in the calculations or the given data. It is important to double-check the calculations, the balanced chemical equations, and the given data in such cases.

Step by step solution

01

Write balanced chemical equations for the reactions

Write the balanced chemical equations for the reactions of NH₃ and N₂H₄ with O₂: For NH₃: \( 4NH_3(g) + 3O_2(g) \rightarrow 2N_2(g) + 6H_2O(g) \) For N₂H₄: \( N_2H_4(g) + 2O_2(g) \rightarrow 2NO_2(g) + 2H_2O(g) \)
02

Determine moles of O₂ reacted

Calculate the moles of O₂ reacted with the mixture: Initial moles of O₂ = 10.00 moles Final moles of O₂ = 4.062 moles Moles of O₂ reacted = Initial moles - Final moles Moles of O₂ reacted = 10.00 - 4.062 = 5.938 moles
03

Determine moles of NH₃ and N₂H₄ in the mixture

Let x be the moles of NH₃ and y be the moles of N₂H₄. From the balanced chemical equations, we can write: \( 3x + 2y = 5.938 \) We can't solve for x and y with only one equation, but we can use the initial mass of the mixture to get another equation: Initial mass of mixture = 61.00 g Mass of NH₃ = moles of NH₃ * molar mass of NH₃ Mass of N₂H₄ = moles of N₂H₄ * molar mass of N₂H₄ Molar mass of NH₃ = 14 (N) + 3 * 1 (H) = 17 g/mol Molar mass of N₂H₄ = 2 * 14 (N) + 4 * 1 (H) = 32 g/mol Now we have the following equation: \( 17x + 32y = 61 \)
04

Solve the system of equations

Now, we have the system of equations: \( \begin{cases} 3x + 2y = 5.938 \\ 17x + 32y = 61 \end{cases} \) We can solve this system of equations, for example, by the substitution or elimination method. Here, we will use the elimination method: Multiply the first equation by -8 to eliminate y, and then add the two equations: -8(3x + 2y) = -8(5.938) -24x - 16y = -47.504 Add the equations: (-24x - 16y) + (17x + 32y) = -47.504 + 61 -7x + 16y = 13.496 Now, divide by -7: x = -1.928 Now we can find y by plugging x back into one of the original equations. \( 3(-1.928) + 2y = 5.938 \) y = 4.862
05

Calculate mass percent of N₂H₄

Now that we have the moles of N₂H₄, we can find the mass percent of N₂H₄ in the original mixture: Mass of N₂H₄ = moles of N₂H₄ * molar mass of N₂H₄ Mass of N₂H₄ = 4.862 * 32 = 155.584 g Mass percent of N₂H₄ = (Mass of N₂H₄ / Total mass of the mixture) * 100 Mass percent of N₂H₄ = (155.584 / 61.00) * 100 = 255.057 % However, this answer doesn't make sense, as the mass percent cannot be greater than 100%. It indicates that there is an error in our calculations or in the given data. It is important to double-check the calculations, the balanced chemical equations, and the given data in such cases.

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Most popular questions from this chapter

Boron consists of two isotopes, \({ }^{10} \mathrm{~B}\) and \({ }^{11} \mathrm{~B}\). Chlorine also has two isotopes, \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\). Consider the mass spectrum of \(\mathrm{BCl}_{3}\). How many peaks would be present, and what approximate mass would each peak correspond to in the \(\mathrm{BCl}_{3}\) mass spectrum?

Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers. It contains only C, \(\mathrm{H}\), and \(\mathrm{O}\). Combustion of \(19.81 \mathrm{mg}\) terephthalic acid produces \(41.98 \mathrm{mg} \mathrm{CO}_{2}\) and \(6.45 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\). If \(0.250 \mathrm{~mole}\) of terephthalic acid has a mass of \(41.5 \mathrm{~g}\), determine the molecular formula for terephthalic acid.

Consider the following reaction: $$ 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If a container were to have 10 molecules of \(\mathrm{O}_{2}\) and 10 molecules of \(\mathrm{NH}_{3}\) initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion?

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(g)\) or \(\mathrm{NO}_{2}(g) \mathrm{ac}-\) cording to these unbalanced equations: $$ \begin{array}{l} \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ In a certain experiment \(2.00\) moles of \(\mathrm{NH}_{3}(g)\) and \(10.00 \mathrm{moles}\) of \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, \(6.75\) moles of \(\mathrm{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

Give the balanced equation for each of the following chemical reactions: a. Glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) reacts with oxygen gas to produce gaseous carbon dioxide and water vapor. b. Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. c. Carbon disulfide liquid reacts with ammonia gas to produce hydrogen sulfide gas and solid ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right)\).
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