Question

A piece of sodium metal reacts completely with water as follows: $$ 2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g) $$ 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 \(25^{\circ} \mathrm{C}=0.0313\) atm. \()\)

Short answer

0.46 grams of sodium were used.

Step-by-step solution

Calculate Partial Pressure of Hydrogen Gas

The total pressure is 1.00 atm, which includes the pressure of water vapor. To find the pressure of the \( H_2 \) gas, subtract the vapor pressure of water from the total pressure. \[ P_{H_2} = P_{total} - P_{water} = 1.00\,\text{atm} - 0.0313\,\text{atm} = 0.9687\,\text{atm} \]

Use Ideal Gas Law to Find Moles of Hydrogen Gas

Use the ideal gas law formula \( PV = nRT \) to calculate the number of moles of \( H_2 \):* \( P = 0.9687 \) atm* \( V = 246\, \text{mL} = 0.246\,\text{L} \)* \( R = 0.0821\,\text{L}\cdot\text{atm}/\text{K}/\text{mol} \)* \( T = 298\,\text{K} \)Calculate the number of moles \( n \):\[ n = \frac{PV}{RT} = \frac{0.9687 \times 0.246}{0.0821 \times 298} = 0.0100\, \text{mol} \]

Relate Moles of Hydrogen to Moles of Sodium

From the balanced equation, 2 moles of \( \mathrm{Na} \) produce 1 mole of \( \mathrm{H}_2 \). So, the moles of \( \mathrm{Na} \) is double the moles of \( \mathrm{H}_2 \).\[ n_{\mathrm{Na}} = 2 \times n_{\mathrm{H}_2} = 2 \times 0.0100 = 0.0200\, \text{mol} \]

Calculate Mass of Sodium

The molar mass of sodium (\( \mathrm{Na} \)) is \( 22.99\,\text{g/mol}\). Use the moles calculated in Step 3 to find the mass.\[ \text{Mass} = n \times \text{molar mass} = 0.0200 \times 22.99 = 0.460\, \text{g} \]

Key concepts

Ideal Gas Law

The Ideal Gas Law is a key concept used in chemistry to relate the pressure, volume, temperature, and number of moles of a gas. The formula is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.

In the context of the exercise, this law is used to calculate the number of moles of hydrogen gas produced during the reaction. Knowing the partial pressure of the hydrogen gas, along with the volume and temperature at which it is collected, allows us to determine the number of moles of hydrogen. This is crucial for connecting the amount of hydrogen gas to the amount of sodium used in the reaction.

Partial Pressure

In gas collections, often the gas of interest is mixed with other gases. Partial pressure is the pressure that a gas would exert if it alone occupied the entire volume. In the problem, the total pressure includes the vapor pressure of water, which is subtracted from the total to find the pressure of the hydrogen gas alone.

Understanding this concept is important because it allows us to accurately calculate the properties of just the hydrogen. By knowing the partial pressure of hydrogen, the Ideal Gas Law can be applied appropriately to find the moles of hydrogen gas specifically, ensuring any calculations are not skewed by the presence of other gases.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, and are balanced by a chemical equation that shows the stoichiometric relationships between the different species involved. In the given reaction, sodium reacts with water to produce sodium hydroxide and hydrogen gas.

The balanced equation is: \[2 \mathrm{Na}(s) + 2 \mathrm{H}_2\mathrm{O}(l) \rightarrow 2 \mathrm{NaOH}(aq) + \mathrm{H}_2(g)\]

By analyzing this equation, we see that two moles of sodium react to produce one mole of hydrogen gas, which directly informs our calculations of how much sodium is used when a certain amount of hydrogen gas is formed. Stoichiometry is used here to relate the amount of hydrogen gas produced back to the sodium consumed.

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It acts as a bridge between the moles of a substance and its mass, making it a fundamental concept in stoichiometric calculations.

For sodium, the molar mass is \(22.99 \, \text{g/mol}\). Once we calculate the moles of sodium involved in the reaction using the stoichiometric relationships from the balanced equation, we can multiply the moles of sodium by its molar mass to find the mass of sodium that reacted. This final step takes us from the theoretical realm of moles to the practical measurement of grams, providing a complete answer to the exercise.