Problem 93 - Chemistry and Chemical Reactivity

Question

A You have \(1.249 \mathrm{g}\) of a mixture of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) You find that \(12.0 \mathrm{mL}\) of \(1.50 \mathrm{M} \mathrm{HCl}\) is required to convert the sample completely to \(\mathrm{NaCl}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{CO}_{2}\) \(\mathrm{NaHCO}_{3}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) $$ \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow\) $$ 2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) $$ What volume of \(\mathrm{CO}_{2}\) is evolved at \(745 \mathrm{mm}\) Hg and \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The volume of \( \mathrm{CO}_{2} \) evolved is approximately 0.335 liters.

Step 1: Determine the Moles of HCl Used

First, we need to calculate the number of moles of hydrochloric acid (HCl) used in the reaction. We are given that the concentration of HCl is 1.50 M (moles per liter) and the volume used is 12.0 mL. To find the moles, use the formula: \[ \text{moles of HCl} = \text{Molarity} \times \text{Volume in Liters} \]Since the volume should be in liters, convert 12.0 mL to liters:\[ 12.0 \text{ mL} = 0.012 \text{ L} \]Now calculate the moles of HCl:\[ \text{moles of HCl} = 1.50 \text{ M} \times 0.012 \text{ L} = 0.018 \text{ moles} \]

Step 2: Set Up the Reaction Stoichiometry

There are two reactions to consider:1. \( \mathrm{NaHCO}_{3} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H}_{2}\mathrm{O} + \mathrm{CO}_{2} \)2. \( \mathrm{Na}_{2}\mathrm{CO}_{3} + 2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl} + \mathrm{H}_{2}\mathrm{O} + \mathrm{CO}_{2} \)In the first reaction, 1 mole of \(\mathrm{HCl}\) produces 1 mole of \(\mathrm{CO}_{2}\), while in the second reaction, 2 moles of \(\mathrm{HCl}\) are needed to produce 1 mole of \(\mathrm{CO}_{2}\).

Step 3: Solve System of Equations for Moles

Let \(x\) be the moles of \(\mathrm{NaHCO}_{3}\) and \(y\) be the moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\). We have two equations based on the stoichiometry:1. \( x + 2y = 0.018 \) (Total moles of HCl)2. \( x M(\mathrm{NaHCO}_{3}) + y M(\mathrm{Na}_{2}\mathrm{CO}_{3}) = 1.249 \) (Total mass of mixture)Where \( M(\mathrm{NaHCO}_{3}) = 84 \text{ g/mol} \) and \( M(\mathrm{Na}_{2}\mathrm{CO}_{3}) = 106 \text{ g/mol} \).

Step 4: Express Moles in Terms of HCl and Mass

Rewriting the mass equation:\[ 84x + 106y = 1.249 \]Substitute \( x = 0.018 - 2y \) into the mass equation:\[ 84(0.018 - 2y) + 106y = 1.249 \]Solve for \(y\):\[ 84(0.018) - 168y + 106y = 1.249 \]\[ 1.512 - 62y = 1.249 \]\[ 62y = 0.263 \]\[ y = \frac{0.263}{62} \approx 0.00424 \text{ moles} \]

Step 5: Calculate Moles of CO2 Evolved

Use the moles of \( x \) and \( y \) to find the total moles of \( \mathrm{CO}_{2} \):Given \( x = 0.018 - 2y \):\[ x = 0.018 - 2(0.00424) = 0.018 - 0.00848 = 0.00952 \text{ moles} \]CO2 from \( \mathrm{NaHCO}_{3} \):\( 0.00952 \) moles.CO2 from \( \mathrm{Na}_{2}\mathrm{CO}_{3} \):\( 0.00424 \) moles.Total moles of \( \mathrm{CO}_{2} \):\( 0.00952 + 0.00424 = 0.01376 \text{ moles} \).

Step 6: Calculate Volume of CO2 at Given Conditions

Use the ideal gas law to find the volume of \( \mathrm{CO}_{2} \):\[ PV = nRT \]Where:- \( P = 745 \text{ mm Hg} = 745 \times \frac{1}{760} \text{ atm} \) - \( V = ? \text{ L} \)- \( n = 0.01376 \text{ moles} \)- \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \)- \( T = 25 + 273 = 298 \text{ K} \)Solve for \( V \):\[ 745/760 \times V = 0.01376 \times 0.0821 \times 298 \]\[ V = \frac{0.01376 \times 0.0821 \times 298}{745/760} \approx 0.335 \text{ L} \]

Step 7: Final Solution

Therefore, the volume of \( \mathrm{CO}_{2} \) evolved is approximately 0.335 liters.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is essential in calculating the volume of gases under various conditions. It's given by the equation \[ PV = nRT \]where:

• \( P \) is the pressure in atmospheres.
• \( V \) is the volume in liters.
• \( n \) is the number of moles.
• \( R \) is the ideal gas constant \( (0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}) \).
• \( T \) is the temperature in Kelvin.

To calculate the volume of \( \mathrm{CO}_2 \) gas produced in the reaction, you need to know these variables. Convert pressure from mm Hg to atm using the conversion \( 1 \text{ atm} = 760 \text{ mm Hg} \). Temperature is converted from Celsius to Kelvin by adding 273. Using these conversions and the moles calculated from stoichiometry, you can easily solve for volume.

Molarity

Molarity is a measure of the concentration of a solute in a solution. It's expressed as moles of solute per liter of solution (M). Understanding molarity is crucial in stoichiometry as it allows you to find the number of moles from a given volume and concentration:\[ \text{moles} = \text{Molarity} \times \text{Volume} \]In the original problem, \(1.50 \text{ M} \) \( \mathrm{HCl} \) is used. By converting the volume from mL to L, you can compute the number of moles of \( \mathrm{HCl} \) involved in the reaction. This helps in determining how much of each reactant is needed and aids in solving the stoichiometry equations.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In this context, the reactions between \( \mathrm{NaHCO}_3 \) and \( \mathrm{Na}_2\mathrm{CO}_3 \) with \( \mathrm{HCl} \) produce \( \mathrm{NaCl}, \mathrm{H}_2\mathrm{O}, \) and \( \mathrm{CO}_2 \). Each reaction requires specific stoichiometric calculations:

• \( \mathrm{NaHCO}_3 + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \)
• \( \mathrm{Na}_2\mathrm{CO}_3 + 2\mathrm{HCl} \rightarrow 2\mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \)

Balancing chemical equations ensures that the same number of atoms of each element appears on both sides, adhering to the Law of Conservation of Mass.

Reaction Stoichiometry

Reaction stoichiometry involves using balanced chemical equations to calculate the amounts of reactants and products. In this exercise, we used stoichiometry to determine how much \( \mathrm{HCl} \) reacts with \( \mathrm{NaHCO}_3 \) and \( \mathrm{Na}_2\mathrm{CO}_3 \). We formed two equations based on the moles of \( \mathrm{HCl} \) and the mass of the mixture:

• \( x + 2y = 0.018 \)
• \( 84x + 106y = 1.249 \)

Solving these equations gives the moles of each compound, allowing you to find the total moles of \( \mathrm{CO}_2 \) produced. This is crucial for further calculations with the Ideal Gas Law to find the gas volume.