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Chapter 3: Problem 122

Consider the following reaction: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ 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?

Short Answer

Expert verified
The limiting reactant in the given reaction is O鈧�. When the reaction goes to completion, there will be 8 molecules of NO, 12 molecules of H鈧侽, and 2 remaining molecules of NH鈧�. So, the total number of molecules present in the container will be 22.

Step by step solution

01

Identify the limiting reactant

To identify the limiting reactant, compare the mole ratios of the given reactants to the balanced equation. Using the balanced equation: 4 NH鈧� + 5 O鈧� 鈫� 4 NO + 6 H鈧侽 We have 10 molecules of NH鈧� and 10 molecules of O鈧�. To find out which reactant will run out first, let's divide the number of molecules by their respective coefficients from the balanced equation. For NH鈧�: \( \frac{10}{4} \) = 2.5 For O鈧�: \( \frac{10}{5} \) = 2 Since the value for O鈧� is smaller, it will be the limiting reactant and will determine the amount of products formed.
02

Determine the moles of products formed

Using the mole ratios in the balanced equation, we will calculate the amount of NO and H鈧侽 produced when all the O鈧� molecules react: For NO The mole ratio of O鈧� to NO is \( \frac{5}{4} \). So, using the limiting reactant O鈧�, we can calculate the number of NO molecules formed: \(NO\, molecules = \frac{4}{5} \times 10 = 8\, molecules\) For H鈧侽 The mole ratio of O鈧� to H鈧侽 is \( \frac{5}{6} \). So, using the limiting reactant O鈧�, we can calculate the amount of H鈧侽 molecules formed: \(H_2O\, molecules = \frac{6}{5} \times 10 = 12\, molecules\)
03

Determine the remaining moles of NH鈧�

Now, we need to determine how many NH鈧� molecules are left after the reaction goes to completion. To do this, we will use the mole ratio of NH鈧� to O鈧� from the balanced equation, which is \( \frac{4}{5} \): \(Used\, NH_3\, molecules = \frac{4}{5} \times 10 = 8\, molecules\) Thus, the remaining NH鈧� molecules: \(Remaining\, NH_3\, molecules = 10 - 8 = 2\, molecules\)
04

Calculate the total number of molecules present after the reaction

Finally, we can calculate the total number of molecules present in the container after the reaction goes to completion by adding the moles of the remaining reactant (NH鈧�) and the products formed (NO and H鈧侽): Total molecules = Remaining NH鈧� molecules + NO molecules + H鈧侽 molecules Total molecules = 2 + 8 + 12 Total molecules = 22 Therefore, there will be 22 total molecules present in the container after the reaction goes to completion.

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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 helps us understand the relationships between the amounts of reactants and products in chemical reactions. It is all about the quantitative relationships.
These relationships are derived from the balanced chemical equation, giving us the tools to calculate how much of each substance is needed or produced.

By using stoichiometry:
  • We can identify limiting reactants, which are the reactants that will be consumed first in the reaction, therefore determining the amount of products formed.
  • We can predict the amounts of products from given quantities of reactants using mole ratios derived from the balanced equation.
  • It ensures that mass is conserved in chemical reactions, as we can track the number of moles and subsequently the number of molecules for each substance involved.
Understanding stoichiometry is key for anyone working in chemistry fields, as it is a cornerstone for predicting reaction outcomes and designing experiments.
Balanced Chemical Equation
A balanced chemical equation provides a recipe for a chemical reaction. It shows the reactants (starting materials) and products (end materials), as well as their respective quantities. The coefficients in front of each substance demonstrate how many units, such as atoms or molecules, are involved.

For instance, the chemical equation in our exercise:
  • \[4 \mathrm{NH}_3(g) + 5 \mathrm{O}_2(g) \rightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_2O(g)\]
denotes that 4 molecules of \( \mathrm{NH}_3 \) react with 5 molecules of \( \mathrm{O}_2 \) to produce 4 molecules of \( \mathrm{NO} \) and 6 molecules of \( \mathrm{H}_2O \).

Balancing equations is crucial because it ensures the law of conservation of mass is satisfied in the reaction.
  • The total number of each type of atom must be the same before and after the reaction.
  • It helps in calculating the reactants and products using stoichiometry, as you need the correct mole ratios to perform these calculations.
  • It provides a clear and concise way to communicate chemical changes.
Mole Ratios
Mole ratios are the heart of stoichiometry and chemical calculations. They are derived from the coefficients of a balanced chemical equation and tell us the relative quantities of reactants and products in a chemical reaction.

To find a mole ratio, simply compare the coefficients from the balanced equation. In the given exercise:
  • The mole ratio of \( \mathrm{O}_2 \) to \( \mathrm{NH}_3 \) is \( \frac{5}{4} \).
  • The mole ratio of \( \mathrm{O}_2 \) to \( \mathrm{NO} \) is \( \frac{5}{4} \).
  • The mole ratio of \( \mathrm{O}_2 \) to \( \mathrm{H}_2O \) is \( \frac{5}{6} \).

Mole ratios are essential for:
  • Determining the limiting reactant, which is the reactant that will be completely consumed first in the reaction, limiting the amount of product formed.
  • Calculating the amounts of other substances needed or produced in a reaction based on the given or required quantity of one substance.
  • Converting between moles and grams of different substances using molar mass.
With mole ratios, chemists can accurately predict the outcomes of chemical reactions, whether in laboratories or industrial processes.

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Most popular questions from this chapter

With the advent of techniques such as scanning tunneling microscopy, it is now possible to "write" with individual atoms by manipulating and arranging atoms on an atomic surface. a. If an image is prepared by manipulating iron atoms and their total mass is \(1.05 \times 10^{-20} \mathrm{g},\) what number of iron atoms were used? b. If the image is prepared on a platinum surface that is exactly 20 platinum atoms high and 14 platinum atoms wide, what is the mass (grams) of the atomic surface? c. If the atomic surface were changed to ruthenium atoms and the same surface mass as determined in part b is used, what number of ruthenium atoms is needed to construct the surface?

Consider the following unbalanced chemical equation for the combustion of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right) :\) $$ \mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If 20.4 g of pentane are burned in excess oxygen, what mass of water can be produced, assuming 100\(\%\) yield?

The most common form of nylon (nylon-6) is 63.68\(\%\) carbon, 12.38\(\%\) nitrogen, 9.80\(\%\) hydrogen, and 14.14\(\%\) oxygen. Calculate the empirical formula for nylon-6.

What mass of sodium hydroxide has the same number of oxygen atoms as 100.0 \(\mathrm{g}\) of ammonium carbonate?

Arrange the following substances in order of increasing mass percent of nitrogen. \(\begin{array}{ll}{\text { a. } \mathrm{NO}} & {\text { c. } \mathrm{NH}_{3}} \\\ {\text { b. } \mathrm{N}_{2} \mathrm{O}} & {\text { d. SNH }}\end{array}\)
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