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Chapter 6: Problem 115

Hydrogen is a possible future fuel. However, elemental hydrogen is rare, so it must be obtained from a hydrogen-containing compound such as water. If hydrogen were obtained from water, how much hydrogen, in grams, could be obtained from 1.0 L of water? (density of water = 1.0 g>cm3 )

Short Answer

Expert verified
112 grams of hydrogen can be obtained from 1.0 L of water.

Step by step solution

01

Calculate the mass of water

Since the density of water is 1.0 g/cm鲁, and we have 1.0 L of water, we first need to convert the volume from liters to cubic centimeters (cm鲁). There are 1000 cm鲁 in 1 L, so the mass of 1.0 L of water is 1.0 L * 1000 cm鲁/L * 1.0 g/cm鲁 = 1000 g of water.
02

Determine the mass of hydrogen in water

Water (H2O) has two hydrogen atoms and one oxygen atom. The molar mass of hydrogen is approximately 1.008 g/mol and for oxygen, it is approximately 16.00 g/mol. The molar mass of water is 2(1.008 g/mol) + 16.00 g/mol = 18.016 g/mol. Since hydrogen makes up 2.016 g/mol of that mass, the fraction of the mass that is hydrogen is 2.016 g/mol / 18.016 g/mol.
03

Calculate the mass of hydrogen that can be obtained

To find the mass of hydrogen that can be obtained from the water, multiply the total mass of water (1000 g) by the fraction of the mass that is hydrogen (2.016 g/mol / 18.016 g/mol). This gives us 1000 g * (2.016 g/mol / 18.016 g/mol) = 112 g (rounded to three significant figures).

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Key Concepts

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

Hydrogen Production from Water
The process of extracting hydrogen from water is an essential science today given our quest for clean, sustainable energy sources. Hydrogen, often referred to as a clean fuel, can be produced through various methods, one of the simplest being electrolysis of water. This method involves passing an electric current through water to separate the hydrogen and oxygen atoms.

Due to its composition, one molecule of water (H2O) contains two hydrogen atoms bonded to one oxygen atom. When electric current is applied, water molecules are split to form hydrogen gas (H2) and oxygen gas (O2). This process not only promises a supply of hydrogen but does so without direct emission of pollutants or greenhouse gases if the electricity used is from a renewable source.
Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships of the substances consumed and produced in chemical reactions. In the context of our hydrogen extraction problem, stoichiometry allows us to calculate the exact amount of hydrogen that can be produced from a known quantity of water.

Understanding the stoichiometry of water, we know that it is comprised of two parts hydrogen to one part oxygen. By using the coefficients of a balanced chemical reaction, which in the case of water's decomposition is 2H2O 鈫� 2H2 + O2, it becomes clear that from two molecules of water, we can obtain two molecules of hydrogen and one molecule of oxygen. This fixed ratio is crucial for predicting the amount of hydrogen gas that will be produced from a given amount of water.
Molar Mass Calculation
The molar mass of a substance is the mass in grams of one mole of that substance. It's a fundamental concept for chemists because it serves as a bridge between the atomic world and the macroscopic world we live in. To calculate the molar mass, we use the atomic masses of the elements comprising the substance, as listed in the periodic table, and add them according to their ratio in a molecule.

In the original problem, the molar mass of hydrogen is around 1.008 g/mol, and oxygen is approximately 16.00 g/mol. Therefore, for water, which has two hydrogen atoms and one oxygen atom (H2O), the molar mass is the sum of twice the atomic mass of hydrogen and the atomic mass of oxygen. This calculation is critical when determining the mass of hydrogen that can be acquired from a specific amount of water, as we need to know the proportion of hydrogen within water's total mass.

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Most popular questions from this chapter

You can use the concepts in this chapter to obtain an estimate of the number of atoms in the universe. These steps will guide you through this calculation. (a) Begin by calculating the number of atoms in the sun. Assume that the sun is pure hydrogen with a density of 1.4 g>cm3 . The radius of the sun is 7 * 108 m, and the volume of a sphere is V = 4 3pr3 . (b) The sun is an average-sized star, and stars are believed to compose most of the mass of the visible universe (planets are so small they can be ignored), so we can estimate the number of atoms in a galaxy by assuming that every star in the galaxy has the same number of atoms as our sun. The Milky Way galaxy is believed to contain 1 * 1011 stars. Use your answer from part a to calculate the number of atoms in the Milky Way galaxy (c) Astronomers estimate that the universe contains approximately 1 * 1011 galaxies. If each of these galaxies contains the same number of atoms as the Milky Way galaxy, what is the total number of atoms in the universe?

Calculate the mass percent composition of O in each compound. (a) calcium nitrate (b) iron(II) sulfate (c) carbon dioxide

You can use mass percent composition as a conversion factor between grams of a constituent element and grams of the compound. Write the conversion factor (including units) inherent in each mass percent composition. (a) Water is \(11.19 \%\) hydrogen by mass. (b) Fructose, also known as fruit sugar, is \(53.29 \%\) oxygen by mass. (c) Octane, a component of gasoline, is \(84.12 \%\) carbon by mass. (d) Ethanol, the alcohol in alcoholic beverages, is \(52.14 \%\) carbon by mass.

A \(1.45-g\) sample of phosphorus bums in air and forms \(2.57 \mathrm{~g}\) of a phosphorus oxide. Calculate the empirical formula of the oxide. (Hint: Determine the mass of oxygen in the \(2.57 \mathrm{~g}\) of phosphorus oxide by determining the difference in mass before and after the phosphorus bums in air.)

What is the mathematical formula for calculating mass percent composition from a chemical formula?
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