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A mixture of \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2}(g)\) reacts in a closed container to form ammonia, \(\mathrm{NH}_{3}(g) .\) The reaction ceases before either reactant has been totally consumed. At this stage \(3.0 \mathrm{mol} \mathrm{N}_{2}, 3.0 \mathrm{mol} \mathrm{H}_{2},\) and 3.0 \(\mathrm{mol} \mathrm{NH}_{3}\) are present. How many moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) were present originally?

Short Answer

Expert verified
There were originally 4.5 moles of \(N_{2}\) and 7.5 moles of \(H_{2}\) present in the container.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the formation of ammonia is: \[N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)\]
02

Calculate moles of N2 and H2 consumed in the reaction

As 3.0 mol NH3 is formed, we can use stoichiometry to determine the moles of N2 and H2 consumed in the reaction. \[\frac{\text{moles of } N_{2}}{\text{moles of } H_{2}} = \frac{1}{3}\] From the balanced equation, we know that \(1 \ mol \ of \ N_{2} \) reacts with \(3 \ mol \ of \ H_{2}\) to produce \(2 \ mol \ of \ NH_{3}\). Therefore, \[\frac{x}{3x} = \frac{1}{3}\] Since we have 3.0 mol of NH3 at equilibrium, we can use stoichiometry to find the moles of N2 and H2 that reacted. \[\frac{x}{3x} = \frac{\text{moles of } N_{2}}{\text{moles of } H_{2}} = \frac{1}{3} \Rightarrow x = 1.5\] So, 1.5 mol of N2 and 4.5 mol of H2 were consumed in the reaction.
03

Determine the original moles

Now that we know how many moles of N2 and H2 were consumed, we can find the original moles when the reaction reaches equilibrium: Original moles of N2 = Moles of N2 at equilibrium + Moles of N2 consumed in the reaction Original moles of N2 = 3.0 mol + 1.5 mol = 4.5 mol Original moles of H2 = Moles of H2 at equilibrium + Moles of H2 consumed in the reaction Original moles of H2 = 3.0 mol + 4.5 mol = 7.5 mol So, there were originally 4.5 moles of N2 and 7.5 moles of H2 present in the container.

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Key Concepts

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

Stoichiometry
Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships of reactants and products in chemical reactions. It tells us how to proportionally mix substances to form certain compounds. In essence, stoichiometry allows chemists to predict how much of a substance is needed or produced in a reaction.

When dealing with stoichiometry, it is crucial to understand the mole concept. A mole is a quantity that measures the number of particles, typically atoms or molecules, in a given sample. This concept is critical because stoichiometry uses moles to relate quantities of different chemicals.

In the given exercise about ammonia synthesis, stoichiometry is used to determine how many moles of nitrogen (\( N_2 \)) and hydrogen (\( H_2 \)) were originally present before the reaction reached equilibrium. This is achieved by understanding the ratios in which reactants combine, as described by the balanced chemical equation. These stoichiometric calculations are the foundation for determining the amounts of reactants consumed and products formed during a reaction.
Balanced Chemical Equation
A balanced chemical equation is essential for representing chemical reactions. It shows the reactants and products along with their relative amounts in moles. This balancing reflects the law of conservation of mass, which dictates that matter cannot be created or destroyed in a chemical reaction.

To balance an equation, one must ensure that the number of each type of atom on the reactant side is equal to the number on the product side. For the ammonia synthesis reaction, the balanced equation is:\[N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)\]

Here, one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia. This equation provides the stoichiometric ratio necessary to calculate how many moles of each reactant are consumed or produced. In the exercise, this balance helps to find out that 1.5 moles of nitrogen and 4.5 moles of hydrogen are consumed to form 3 moles of ammonia during the reaction.
  • Equal Atoms: Ensure equal numbers of atoms for each element.
  • Whole Number Coefficients: Use whole numbers to represent mole proportions.
  • Verify Balancing: Double-check that each element's count is balanced.
Ammonia Synthesis
Ammonia synthesis is a critical industrial process, commonly known as the Haber-Bosch process. It involves combining nitrogen (\( N_2 \)) from the air with hydrogen (\( H_2 \)) under high temperature and pressure, using a catalyst to form ammonia (\( NH_3 \)). This is a vital process as ammonia is a key ingredient in fertilizers, essential for global agriculture.
During the synthesis, nitrogen and hydrogen gases react according to the balanced chemical equation. This reaction occurs in a closed system, and chemical equilibrium is reached when the rate of formation of ammonia equals the rate of decomposition back into nitrogen and hydrogen.
In our exercise, we focus on ammonia synthesis and the stoichiometric relationship between reactants and products. Even though the reaction stops before either reactant is fully consumed, it's crucial to determine the initial amounts based on reactions at equilibrium. Understanding this process helps in efficiently managing reactants and predicting outcomes in industrial applications.

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Most popular questions from this chapter

Hydrogen cyanide, HCN, is a poisonous gas. The lethal dose is approximately 300 \(\mathrm{mg}\) HCN per kilogram of air when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring \(12 \times 15 \times 8.0 \mathrm{ft}\) . The density of air at \(26^{\circ} \mathrm{C}\) is 0.00118 \(\mathrm{g} / \mathrm{cm}^{3} .\) (b) If the HCN is formed by reaction of \(\mathrm{NaCN}\) with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4}\) what mass of NaCN gives the lethal dose in the room? $$ 2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g) $$ (c) HCN forms when synthetic fibers containing Orlon or Acrilan burn. Acrilan has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN},\) so HCN is 50.9\(\%\) of the formula by mass. A rug measures \(12 \times 15 \mathrm{ft}\) and contains 30 oz of Acrilan fibers per square yard of carpet. If the rug burns, will a lethal dose of HCN be generated in the room? Assume that the yield of HCN from the fibers is 20\(\%\) and that the carpet is 50\(\%\) consumed.

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