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Verify that the following data confirm the law of equivalent proportions: Nitrogen and oxygen react with hydrogen to form ammonia and water, respectively: \(4.66 \mathrm{~g}\) of nitrogen is required for every gram of hydrogen in ammonia, and \(8 \mathrm{~g}\) of oxygen for every gram of hydrogen in water. Nitrogen plus oxygen yields NO. Here, \(14 \mathrm{~g}\) of nitrogen is required for every 16 g of oxygen.

Short Answer

Expert verified
The given data confirms the law of equivalent proportions, as the ratio of nitrogen to hydrogen in ammonia (\(4.66\)) multiplied by the ratio of nitrogen to oxygen in NO (\(\frac{7}{8}\)) is approximately equal to the ratio of oxygen to hydrogen in water (4.0775 ≈ 4).

Step by step solution

01

Calculate the ratio of nitrogen to hydrogen in ammonia

To find the ratio of nitrogen to hydrogen in ammonia, we divide the mass of nitrogen by the mass of hydrogen. Given, for ammonia: 4.66 g of nitrogen required for every gram of hydrogen. Ratio of nitrogen to hydrogen in ammonia = \(\frac{4.66}{1}\) = 4.66
02

Calculate the ratio of oxygen to hydrogen in water

To find the ratio of oxygen to hydrogen in water, we divide the mass of oxygen by the mass of hydrogen. Given, for water: 8 g of oxygen required for every gram of hydrogen. Ratio of oxygen to hydrogen in water = \(\frac{8}{1}\) = 8
03

Calculate the ratio of nitrogen to oxygen in NO

To find the ratio of nitrogen to oxygen in NO, we divide the mass of nitrogen by the mass of oxygen. Given, for NO: 14 g of nitrogen required for every 16 g of oxygen. Ratio of nitrogen to oxygen in NO = \(\frac{14}{16}\) = \(\frac{7}{8}\)
04

Compare the ratios to verify the law of equivalent proportions

According to the law of equivalent proportions, the ratio of nitrogen to hydrogen in ammonia and the ratio of oxygen to hydrogen in water should be related to the ratio of nitrogen to oxygen in NO. Now, let's multiply the ratio of nitrogen to hydrogen in ammonia(\(4.66\)) with the ratio of nitrogen to oxygen in NO(\(\frac{7}{8}\)): \(4.66 \times \frac{7}{8} = \frac{4.66 \times 7}{8} = \frac{32.62}{8} = 4.0775\) Comparing this value with the ratio of oxygen to hydrogen in water: 4.0775 (approx) ≈ 4 The ratio of nitrogen to hydrogen in ammonia when multiplied by the ratio of nitrogen to oxygen in NO is approximately equal to the ratio of oxygen to hydrogen in water, which confirms the law of equivalent proportions.

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Key Concepts

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

Stoichiometry
Stoichiometry is essentially the math behind chemistry. It involves calculating the quantities of reactants and products in a chemical reaction to ensure that the reaction obeys the law of conservation of mass. This law states that in a chemical reaction, matter is not created or destroyed, but transformed from one form to another. Stoichiometry relies on the balanced chemical equation, where the coefficients represent the relative amounts of moles of each substance involved.

Therefore, understanding stoichiometry is crucial for predicting how much of a given reactant is needed to produce a desired amount of product, which has practical implications in fields ranging from industrial manufacturing to pharmaceuticals. A firm grasp of this concept helps explain why the ratio of nitrogen to hydrogen in ammonia is critical for students to conceptualize how these components react on a molecular level.
Chemical Reaction Ratios
Chemical reaction ratios are at the core of understanding how different substances interact in a reaction. These ratios, derived from balanced chemical equations, tell us the exact proportions of reactants that combine and the amounts of products formed as a result. For example, the ratio of nitrogen to hydrogen in the formation of ammonia is a fixed value, illustrating that a specific amount of nitrogen is needed to react with a certain amount of hydrogen to form a set amount of ammonia, without leaving excess reactants.

It's like following a recipe in cooking; you need to measure the ingredients accurately to obtain the expected result. If the ratio is incorrect, the chemical reaction will either be incomplete, or there will be an excess of one component. Using the given ratios of reactants in the exercise aids students in visualizing and quantifying the precise chemical relationship at work.
Chemical Law Validation
The law of equivalent proportions is a fundamental principle in chemistry stating that when elements combine to form more than one compound, the masses of one element which combine with a fixed mass of the other are in simple ratio to one another. In the provided exercise, verifying the law involves using the calculated ratios of elements in different compounds and comparing them to establish this fixed relationship.

By successfully demonstrating that the product of the nitrogen to hydrogen ratio in ammonia and the nitrogen to oxygen ratio in NO closely approximates the oxygen to hydrogen ratio in water, students validate this chemical law. This kind of exercise not only reinforces their understanding of stoichiometric concepts but also imparts confidence in the predictable nature of chemical reactions, which is the fundamental rationale for the systematic study of chemistry.

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Most popular questions from this chapter

Determine the weights of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced on burning \(104 \mathrm{~g}\). of \(\mathrm{C}_{2} \mathrm{H}_{2}\). Molecular weights are \(\mathrm{CO}_{2}=44\) and \(\mathrm{H}_{2} \mathrm{O}=18\). The equation for the reaction is \(2 \mathrm{C}_{2} \mathrm{H}_{2}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)

What volume of ammonia at STP can be obtained when steam is passed over \(4000 \mathrm{~g}\) of calcium cyanamide? The balanced reaction is \(\mathrm{CaCN}_{2}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NH}_{3}+\mathrm{CaCO}_{3}\) (Molecular weight of \(\mathrm{CaCN}_{2}=80, \mathrm{MW}\) of \(\mathrm{NH}_{3}=17 .\) )

How many pounds of air (which is \(23.19 \% \mathrm{O}_{2}\) and \(75.46 \%\) \(\mathrm{N}_{2}\) by weight) would be needed to burn a pound of gasoline by a reaction whereby \(\mathrm{C}_{8} \mathrm{H}_{18}\) reacts with \(\mathrm{O}_{2}\) to form \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ?

What is the mass of 1 liter of carbon monoxide (CO) at standard temperature and pressure (STP).

A chemist has a mixture of \(\mathrm{KClO}_{3}, \mathrm{KHCO}_{3}, \mathrm{~K}_{2} \mathrm{CO}_{3}\), and \(\mathrm{KCl}\). She heats \(1,000 \mathrm{~g}\) of this mixture and notices that the following gases evolve: 18 g of water \(\left(\mathrm{H}_{2} \mathrm{O}\right), 132 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(40 \mathrm{~g}\) of \(\mathrm{O}_{2}\) according to the following reactions: $$ \begin{array}{ll} 2 \mathrm{KC} 1 \mathrm{O}_{3} & \rightarrow 2 \mathrm{KCl}+3 \mathrm{O}_{2} \\\ 2 \mathrm{KHCO}_{3} & \rightarrow \mathrm{K}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{CO}_{2} \\ \mathrm{~K}_{2} \mathrm{CO}_{3} & \rightarrow \mathrm{K}_{2} \mathrm{O}+\mathrm{CO}_{2} \end{array} $$ The \(\mathrm{KCl}\) is inert under these conditions. Assuming complete decomposition, determine the composition of the original mixture.

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