Problem 71
Question
Consider the chemical reaction: $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$ How many liters of hydrogen gas are formed from the complete reaction of \(15.7 \mathrm{~g}\) C? Assume that the hydrogen gas is collected at a pressure of \(1.0 \mathrm{~atm}\) and a temperature of \(355 \mathrm{~K}\).
Step-by-Step Solution
Verified Answer
The volume of hydrogen gas produced is calculated using the Ideal Gas Law with the moles of carbon reacted.
1Step 1: Convert grams of carbon to moles
Use the molecular weight of carbon (C), which is 12.01 g/mol, to convert the mass of carbon to moles. This is done using the formula: moles of C = mass of C / molar mass of C.
2Step 2: Write the balanced chemical equation
The balanced chemical equation is already given:\[\mathrm{C}(s) + \mathrm{H}_{2}\mathrm{O}(g) \longrightarrow \mathrm{CO}(g) + \mathrm{H}_{2}(g)\]This shows a stoichiometry of 1 mole of C producing 1 mole of \(\mathrm{H}_{2}\) gas.
3Step 3: Use the stoichiometry to find moles of \(\mathrm{H}_{2}\)
Since the stoichiometry of the reaction is 1:1 for carbon to hydrogen gas, the moles of hydrogen gas produced is equal to the moles of carbon reacted.
4Step 4: Use the Ideal Gas Law to find the volume of \(\mathrm{H}_{2}\)
Use the Ideal Gas Law, which is \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is moles of gas, \(R\) is the universal gas constant (0.0821 L·atm/mol·K), and \(T\) is the temperature in Kelvin. Solve for \(V\) to find the volume of hydrogen gas produced.
5Step 5: Calculate the volume of \(\mathrm{H}_{2}\)
Plug in the values into the Ideal Gas Law equation: \(V = \frac{nRT}{P}\). Use the moles of \(\mathrm{H}_{2}\) calculated in Step 3, \(R = 0.0821 \frac{L·atm}{mol·K}\), the given temperature (355 K), and the pressure (1.0 atm) to calculate the volume of \(\mathrm{H}_{2}\) gas in liters.
Key Concepts
StoichiometryIdeal Gas LawChemical Reaction QuantificationMolar Mass
Stoichiometry
Stoichiometry is a branch of chemistry that quantifies the relationships between the reactants and products in a chemical reaction. It allows us to predict the amounts of substances consumed and produced during a reaction, based on the balanced chemical equation. In stoichiometry, the mole ratio of the reactants and products, which stems directly from the coefficients in the balanced equation, plays a crucial role.
For the given reaction, \[\mathrm{C}(s) + \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g) + \mathrm{H}_{2}(g)\], the stoichiometry shows a 1:1 ratio between carbon (\mathrm{C}) and hydrogen gas (\mathrm{H}_{2}). Therefore, for each mole of carbon that reacts, one mole of hydrogen gas is produced. This understanding is pivotal for determining the quantities of reactants needed or products formed in any chemical process.
For the given reaction, \[\mathrm{C}(s) + \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g) + \mathrm{H}_{2}(g)\], the stoichiometry shows a 1:1 ratio between carbon (\mathrm{C}) and hydrogen gas (\mathrm{H}_{2}). Therefore, for each mole of carbon that reacts, one mole of hydrogen gas is produced. This understanding is pivotal for determining the quantities of reactants needed or products formed in any chemical process.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas. It is usually expressed as \(PV = nRT\), where \(R\) is the universal gas constant, with a value of 0.0821 L·atm/mol·K. This law is incredibly useful for predicting the behavior of gases under various conditions.
When solving for the volume of a gas, as is necessary in this exercise, we can rearrange the Ideal Gas Law to \(V = \frac{nRT}{P}\). By knowing the amount of gas in moles, the temperature in Kelvin, and the pressure, one can calculate the occupied volume. This is particularly helpful when dealing with gaseous products in chemical reactions, like hydrogen gas (\mathrm{H}_{2}) in the given exercise.
When solving for the volume of a gas, as is necessary in this exercise, we can rearrange the Ideal Gas Law to \(V = \frac{nRT}{P}\). By knowing the amount of gas in moles, the temperature in Kelvin, and the pressure, one can calculate the occupied volume. This is particularly helpful when dealing with gaseous products in chemical reactions, like hydrogen gas (\mathrm{H}_{2}) in the given exercise.
Chemical Reaction Quantification
Quantification in chemical reactions involves measuring or calculating the amount of reactants and products. When given the mass of a reactant, as in this exercise, the first step is to convert the mass to moles using the molar mass of the substance. Once you have moles, you can apply the stoichiometry of the reaction to find the amount of product formed.
With the balanced equation and the Ideal Gas Law at hand, we complete the quantification by calculating the volume the gaseous product will occupy at specific conditions of temperature and pressure. Accurate quantification is essential in both the academic problems and practical applications like industrial chemical synthesis and environmental monitoring.
With the balanced equation and the Ideal Gas Law at hand, we complete the quantification by calculating the volume the gaseous product will occupy at specific conditions of temperature and pressure. Accurate quantification is essential in both the academic problems and practical applications like industrial chemical synthesis and environmental monitoring.
Molar Mass
The molar mass is the weight of one mole of a substance, normally expressed in grams per mole (g/mol). It is a pivotal concept in stoichiometry because it serves as the conversion factor between mass and moles for chemical substances. The molar mass of an element like carbon (C) is taken from the periodic table, where carbon has a molar mass of approximately 12.01 g/mol.
To solve the exercise in question, one would divide the given mass of carbon (15.7 grams) by its molar mass to find the moles of carbon, which then directly corresponds to the moles of hydrogen gas produced due to the 1:1 mole ratio in the reaction. Thus, this simple concept enables the translation from mass to moles, a fundamental step in the analysis of any chemical reaction.
To solve the exercise in question, one would divide the given mass of carbon (15.7 grams) by its molar mass to find the moles of carbon, which then directly corresponds to the moles of hydrogen gas produced due to the 1:1 mole ratio in the reaction. Thus, this simple concept enables the translation from mass to moles, a fundamental step in the analysis of any chemical reaction.
Other exercises in this chapter
Problem 59
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