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Chapter 3: Problem 117

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{~g} \mathrm{~N}_{2}\) and \(5.00 \times 10^{2} \mathrm{~g} \mathrm{H}_{2} ?\) b. What mass of which starting material would remain unreacted?

Short Answer

Expert verified
The maximum mass of ammonia that can be produced is \(1.22\times 10^3~g~NH_{3}\). The mass of the leftover starting material, hydrogen gas, is approximately 283.6 g.

Step by step solution

01

Calculate the moles of the starting materials

First, we need to convert the mass of nitrogen and hydrogen gases into moles. To do this, we will use the molar mass of each gas. The molar mass of N鈧� is 28.02 g/mol and H鈧� is 2.02 g/mol. Nitrogen gas: \( \frac{1.00 \times 10^3 g~N_{2}}{28.02 g~N_{2}/mol~N_{2}} = 35.7 mol~N_{2} \) Hydrogen gas: \( \frac{5.00 \times 10^2 g~H_{2}}{2.02 g~H_{2}/mol~H_{2}} = 247.5 mol~H_{2} \)
02

Determine the limiting reactant

Using the stoichiometry of the balanced equation, we can determine which reactant is the limiting reactant. In this reaction, one mole of N鈧� reacts with three moles of H鈧� to produce two moles of NH鈧�. To find the limiting reactant, we can divide the number of available moles of each reactant by their respective stoichiometric coefficients and find the lowest value. For N鈧�: \( \frac{35.7 mol~N_{2}}{1} = 35.7 \) For H鈧�: \( \frac{247.5 mol~H_{2}}{3} = 82.5 \) Since 35.7 is the lowest value, N鈧� is the limiting reactant.
03

Calculate the maximum mass of ammonia produced

Using the stoichiometry of the balanced equation and the limiting reactant, we can find the moles of ammonia produced. In this reaction, one mole of N鈧� produces two moles of NH鈧�. The molar mass of NH鈧� is 17.03 g/mol. \( Moles~of~NH_{3} = 35.7 mol~N_{2} \times \frac{2~mol~NH_{3}}{1~mol~N_{2}} = 71.4 mol~NH_{3} \) Now convert the moles of ammonia produced into mass: \( Mass~of~NH_{3} = 71.4 mol~NH_{3} \times \frac{17.03 g~NH_{3}}{1 mol~NH_{3}} = 1.22\times 10^3 g~NH_{3} \) The maximum mass of ammonia that can be produced is \(1.22\times 10^3~g~NH_{3}\). #b. Finding the mass of the leftover starting material#
04

Calculate the moles of the leftover starting material

Since N鈧� is the limiting reactant, we need to determine the moles of hydrogen that were not consumed in the reaction. We can use the stoichiometry of the balanced equation again: \( Moles~of~unused~H_{2} = 35.7 mol~N_{2} \times \frac{3~mol~H_{2}}{1~mol~N_{2}} = 107.1 mol~H_{2} \) Next, find the remaining moles of H鈧�: \( Remaining~H_{2} = Initial~H_{2} - Used~H_{2} = 247.5 mol~H_{2} - 107.1 mol~H_{2} = 140.4 mol~H_{2} \)
05

Convert the moles of the leftover starting material into mass

Finally, convert the moles of leftover hydrogen gas into mass: \( Mass~of~leftover~H_{2} = 140.4 mol~H_{2} \times \frac{2.02 g~H_{2}}{1 mol~H_{2}} = 283.6 g~H_{2} \) The mass of the leftover starting material, hydrogen gas, is approximately 283.6 g.

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Key Concepts

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

Limiting Reactant
The limiting reactant in a chemical reaction is the substance that is totally consumed when the reaction is complete. It limits the amount of product that can be formed. To identify the limiting reactant, we compare the mole ratios of the reactants to the balanced chemical equation. In our given exercise,
  • The balanced equation is \[\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)\]
  • One mole of nitrogen (\(\mathrm{N}_2\)) requires three moles of hydrogen (\(\mathrm{H}_2\)).
We calculated the amount of moles for both reactants:
  • \(35.7\) moles of nitrogen and \(247.5\) moles of hydrogen.
By dividing these amounts by their respective coefficients in the balanced equation:
  • Nitrogen: \(\frac{35.7}{1} = 35.7\)
  • Hydrogen: \(\frac{247.5}{3} = 82.5\)
Since nitrogen has the smaller result, it is the limiting reactant. This means ammonia production is capped by the available nitrogen.
Chemical Equations
Chemical equations represent the transformation of reactants into products in a chemical reaction. They are essential because they provide a concise way to describe a chemical reaction. For our reaction, the equation is:\[\mathrm{N}_{2}(g) + 3\mathrm{H}_{2}(g) \rightarrow 2\mathrm{NH}_{3}(g)\]In this balanced chemical equation:
  • One molecule of nitrogen gas (\(\mathrm{N}_2\)) reacts with three molecules of hydrogen gas (\(\mathrm{H}_2\)).
  • This produces two molecules of ammonia (\(\mathrm{NH}_3\)).
Balancing chemical equations ensures that the law of conservation of mass is maintained. This means the same number of each type of atom has to be present on both sides of the equation. Balancing provides the stoichiometric coefficients, critical for calculating reactant and product masses.
Molar Mass
Molar mass is fundamental when converting between grams and moles in chemistry. It is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). This concept allows chemists to calculate how much substance is involved in a reaction.For example, let's examine the components in our ammonia synthesis problem:
  • \(\mathrm{N}_2\): 28.02 g/mol.
  • \(\mathrm{H}_2\): 2.02 g/mol.
  • \(\mathrm{NH}_3\): 17.03 g/mol.
These values help convert the starting gram amount into moles, which we used in stoichiometric calculations to find limiting reactants and predict the amounts of products formed. Understanding molar mass is crucial for accurately quantifying reactions.
Chemical Reactions
Chemical reactions are processes where reactants are transformed into products. They are the foundation of all chemical processes and syntheses. The reaction in our example鈥攕ynthesizing ammonia from nitrogen and hydrogen鈥攊s a classic industrial application known as the Haber process. In chemical reactions:
  • Reactants are the starting materials; in this case, nitrogen and hydrogen gases.
  • Products are what is formed as a result; here, it's ammonia.
Chemical reactions can be categorized by type, such as synthesis, decomposition, single replacement, and double replacement. The one we explored is a synthesis reaction, where simpler substances combine to form a more complex compound. Understanding these basic concepts about reactions is essential in stoichiometry for calculating how much of a product can be created from given amounts of reactants.

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Most popular questions from this chapter

In chemistry, what is meant by the term "mole"? What is the importance of the mole concept?

Consider the following balanced chemical equation: A 1 5B h 3C 1 4D a. Equal masses of A and B are reacted. Complete each of the following with either 鈥淎 is the limiting reactant because ________鈥�; 鈥淏 is the limiting reactant because ________鈥�; or 鈥渨e cannot determine the limiting reactant because ________.鈥� i. If the molar mass of A is greater than the molar mass of B, then ii. If the molar mass of B is greater than the molar mass of A, then b. The products of the reaction are carbon dioxide (C) and water (D). Compound A has a similar molar mass to carbon dioxide. Compound B is a diatomic molecule. Identify compound B, and support your answer. c. Compound A is a hydrocarbon that is 81.71% carbon by mass. Determine its empirical and molecular formulas

Which of the following statements about chemical equations is(are) true? a. When balancing a chemical equation, you can never change the coefficient in front of any chemical formula. b. The coefficients in a balanced chemical equation refer to the number of grams of reactants and products. c. In a chemical equation, the reactants are on the right and the products are on the left. d. When balancing a chemical equation, you can never change the subscripts of any chemical formula. e. In chemical reactions, matter is neither created nor destroyed so a chemical equation must have the same number of atoms on both sides of the equation.

A compound containing only sulfur and nitrogen is \(69.6 \% \mathrm{~S}\) by mass; the molar mass is \(184 \mathrm{~g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?

Ascorbic acid, or vitamin \(\mathrm{C}\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\), is an essential vitamin. It cannot be stored by the body and must be present in the diet. What is the molar mass of ascorbic acid? Vitamin C tablets are taken as a dietary supplement. If a typical tablet contains \(500.0 \mathrm{mg}\) vitamin \(\mathrm{C}\), what amount (moles) and what number of molecules of vitamin C does it contain?
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