Question

A piece of sodium metal undergoes complete reaction with water as follows $$2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(aq)+\mathrm{H}_{2}(g)$$ The hydrogen gas generated is collected over water at \(25.0^{\circ} \mathrm{C}\). The volume of the gas is \(246 \mathrm{~mL}\) measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at \(\left.25^{\circ} \mathrm{C}=0.0313 \mathrm{~atm} .\right)\)

Short answer

To derive the short answer, calculate the moles of Hydrogen first using the corrected pressure, volume, and the ideal gas law. Then, double this value to find the moles of Sodium used. Finally, multiply the moles of Sodium by its molar mass to find the mass of Sodium used in grams.

Step-by-step solution

Determine the moles of Hydrogen produced

Since the volume of the Hydrogen gas produced is given, along with the temperature and pressure, the moles of H2 can be calculated using the Ideal Gas Law, expressed as \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. However, first, the total pressure must be corrected for the vapor pressure of the water. The pressure exerted by the Hydrogen gas should then be: 1.00 atm - 0.0313 atm = 0.68 atm. Convert the temperature to Kelvin (T = 25C = 298K), and the volume to liters (V = 246 mL = 0.246 L). Using the Ideal Gas Law, the moles of H2 (n) can be solved: \(n= PV/R*T\) where R is 0.0821 L.atm/mol.K. This will yield the number of moles of H2.

Apply the Stoichiometry of the Reaction

In the balanced chemical equation, 2 moles of Na result in the production of 1 mole of H2. Therefore, the moles of Sodium consumed in the reaction would be twice the moles of Hydrogen calculated in the previous step.

Convert Moles of Sodium to Grams

The molar mass of Sodium is roughly 23 g/mol. Therefore, by multiplying the number of moles of Sodium with this molar mass, one can determine the mass of Sodium that was used in the reaction.

Key concepts

Ideal Gas Law

Understanding the Ideal Gas Law is fundamental when working with gases, and it is especially useful in stoichiometry calculations involving a gas. The Ideal Gas Law is given by the equation \( PV = nRT \). Here's what each symbol represents:

• \(P\) stands for the pressure of the gas.
• \(V\) is the volume of the gas.
• \(n\) indicates the number of moles of the gas.
• \(R\) is the gas constant, which has a value of 0.0821 L.atm/mol.K for these units.
• \(T\) refers to the temperature in Kelvin.

To apply the Ideal Gas Law in stoichiometry problems, such as the reaction between sodium and water, you need to be sure the gas volume, pressure, and temperature are properly converted into the correct units—liters for volume, atmospheres for pressure, and Kelvin for temperature. Remembering that 1 atm is the standard pressure and R is the ideal gas constant, you can rearrange the Ideal Gas Law to solve for the unknown variable (in most cases, the moles of gas). It is also essential to correct the pressure for any vapor present; otherwise, you might not accurately calculate the moles of gas.

Chemical Reaction with Sodium and Water

The interaction between sodium and water is a classic example of a chemical reaction that students encounter in chemistry. The equation \(2 \text{Na}(s) + 2 \text{H}_2\text{O}(l) \rightarrow 2 \text{NaOH}(aq) + \text{H}_2(g)\) summarizes this exothermic and highly reactive process, in which solid sodium reacts vigorously with water to produce sodium hydroxide and hydrogen gas.

Understanding this type of reaction requires recognizing that stoichiometry—the calculation of reactants and products in chemical reactions—is governed by the coefficients in the balanced chemical equation. These coefficients tell us, for instance, that two moles of sodium react with two moles of water to produce two moles of sodium hydroxide and one mole of hydrogen gas. This kind of knowledge is paramount when trying to determine how much of one reactant (like sodium) is needed to produce a certain amount of product (hydrogen gas, in this case).

Mole-to-Mass Conversion

The mole-to-mass conversion is a critical tool in stoichiometry, which allows chemists to translate the language of atoms and molecules—measured in moles—into something more tangible: mass. A mole is Avogadro's number of particles, which is \(6.022 \times 10^{23}\) entities. To convert from moles to mass, multiply the number of moles by the molar mass of the substance, which is unique for each element or compound.

For sodium (Na), the molar mass is roughly 23 grams per mole. Therefore, if a stoichiometry calculation reveals that you need a certain number of moles of sodium to react (let's say you find you need 2 moles), you would then multiply this amount by sodium's molar mass to find out that you require 46 grams of sodium. This straightforward calculation is the core of the process chemists use to scale reactions from the molecular level to practical, real-world quantities.

Vapor Pressure Correction

Vapor pressure correction is a crucial step in many gas-related stoichiometry problems, particularly when the gas is collected over water, as in the sodium-water reaction. This correction accounts for the partial pressure contributed by water vapor, which can affect the overall pressure of the collected gas.

At any given temperature, water has a characteristic vapor pressure which is the pressure exerted by water molecules in the vapor phase at equilibrium. To find the pressure of the dry gas alone (like hydrogen in our exercise), you subtract the vapor pressure of water from the total pressure measured. In our exercise, the total pressure is given as 1.00 atm; the vapor pressure of water at \(25^{\text{o}}C\) is 0.0313 atm. Thus, the pressure of the hydrogen gas alone is 1.00 atm - 0.0313 atm = 0.9687 atm. This corrected pressure is utilized in the Ideal Gas Law to determine the accurate amount of hydrogen gas produced during the reaction.