Question

If all volumes of gases are measured under the same conditions, what volume of the product will be formed from reaction of \(5.00 \mathrm{~L}\) of \(\mathrm{H}_{2}\) in each of the following reactions, assuming they go to completion? (a) \(\mathrm{S}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (b) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (c) \(\mathrm{C}_{6} \mathrm{H}_{6}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12}(g)\)

Short answer

(a) 5.00 L of H2S. (b) 3.33 L of NH3. (c) 1.67 L of C6H12.

Step-by-step solution

Understanding the Stoichiometry of the Reactions

First, take a look at the stoichiometric coefficients of each reaction. They tell exactly how many moles of reactants react with each other or are produced. (a) In the first reaction, one molecule of S reacts with one molecule of H2 to form one molecule of H2S. This means that each liter of H2 will produce one liter of H2S. (b) In the second reaction, one molecule of N2 reacts with three molecules of H2 to form two molecules of NH3. Thus, each liter of H2 will produce 2/3 liter of NH3. (c) In the third reaction, one molecule of C6H6 reacts with three molecules of H2 to form one molecule of C6H12. Thus, each liter of H2 will produce 1/3 liter of C6H12.

Calculating the Volume of Each Product

Next, use the stoichiometric coefficients of the reactions to calculate the volume of each product after the reaction. (a) For the first reaction, multiply the initial volume of H2 (5.00L) by 1 (coefficient ratio 1/1), to give 5.00L of H2S. (b) For the second reaction, multiply the initial volume of H2 (5.00L) by 2/3 (coefficient ratio 2/3), to give 3.33L of NH3. (c) For the third reaction, multiply the initial volume of H2 (5.00L) by 1/3 (coefficient ratio 1/3), to give 1.67L of C6H12.

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, called products. In our exercise, various gases react together under the same conditions to form different products. Each reaction entails certain elements switching places or combining to form a completely new compound.
Understanding a chemical equation is key to knowing which products and reactants are involved:

• The reactants are on the left side of the equation, and the products are on the right.
• For instance, in the reaction: \( \mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3} \), nitrogen (\( \mathrm{N}_{2} \)) and hydrogen (\( \mathrm{H}_{2} \)) are reactants, and ammonia (\( \mathrm{NH}_{3} \)) is the product.
• Each molecule or atom involved in the reaction is represented by its chemical formula, such as \( \mathrm{H}_{2} \) for hydrogen gas.
• The numbers in front of these formulas, called stoichiometric coefficients, indicate how many molecules are involved.

Gas Volume Calculations

Gas volume calculations are crucial in stoichiometry, especially when these gases participate in reactions under similar conditions. Gases have unique properties that allow their volumes to be compared directly.
According to Avogadro's law, equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules. This concept helps simplify calculations, allowing us to equate the stoichiometric coefficients directly to gas volumes.

• Volume relationships in gas reactions can be determined by the coefficients in a balanced chemical equation.
• In reaction (b) \( \mathrm{N}_{2} + 3\mathrm{H}_{2} \rightarrow 2\mathrm{NH}_{3} \), using stoichiometry, we find that 3 volumes of \( \mathrm{H}_{2} \) produce 2 volumes of \( \mathrm{NH}_{3} \).
• Therefore, if 5.00 L of \( \mathrm{H}_{2} \) is used, the volume of \( \mathrm{NH}_{3} \) produced is \( \frac{2}{3} \times 5.00 \text{ L} = 3.33 \text{ L} \).

Keeping these relationships in mind allows us to predict the volumes of products formed during any gas-phase chemical reaction.

Molar Ratios

The concept of molar ratios is central to stoichiometry and involves the proportions of reactants and products in a chemical reaction. It indicates how much of one substance reacts or forms in relation to another
In our reactions, molar ratios are evident directly from the balanced equation:

• For example, the reaction \( \mathrm{S} + \mathrm{H}_{2} \rightarrow \mathrm{H}_{2}\mathrm{S} \) shows a molar ratio of 1:1 between \( \mathrm{H}_{2} \) and \( \mathrm{H}_{2}\mathrm{S} \).
• This means that one mole, or equivalent volume, of \( \mathrm{H}_{2} \) produces one mole, or volume, of \( \mathrm{H}_{2}\mathrm{S} \).
• Similarly, for \( \mathrm{C}_{6}\mathrm{H}_{6} + 3\mathrm{H}_{2} \rightarrow \mathrm{C}_{6}\mathrm{H}_{12} \), the molar ratio of \( \mathrm{H}_{2} \) to \( \mathrm{C}_{6}\mathrm{H}_{12} \) is 3:1, meaning 3 moles of \( \mathrm{H}_{2} \) yield 1 mole of \( \mathrm{C}_{6}\mathrm{H}_{12} \).

Utilizing these ratios helps to conveniently calculate how much of a product forms or how much reactant is needed, aligning chemical quantities with their real-world applications.