Problem 61 - Chemistry and Chemical Reactivity

Question

What volume of \(0.109 \mathrm{M} \mathrm{HNO}_{3},\) in milliliters, is required to react completely with \(2.50 \mathrm{g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2} ?\) $$2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{s}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})$$

Step-by-Step Solution

Answer

268 mL of \(0.109 \mathrm{M} \mathrm{HNO}_3\) is required.

Step 1: Calculate Moles of Ba(OH)_2

First, find the molar mass of \(\mathrm{Ba(OH)}_2\). The atomic masses are Ba = 137.33 g/mol, O = 16.00 g/mol, and H = 1.01 g/mol. Therefore, the molar mass of \(\mathrm{Ba(OH)}_2\) is: \[137.33 + 2(16.00 + 1.01) = 171.35 \text{ g/mol.}\] Now calculate the moles of \(\mathrm{Ba(OH)_2}\) using the given mass: \[\text{Moles of } \mathrm{Ba(OH)}_2 = \frac{2.50 \text{ g}}{171.35 \text{ g/mol}} = 0.0146 \text{ moles.}\]

Step 2: Use Stoichiometry to Find Moles of HNO_3

According to the balanced chemical equation, 1 mole of \(\mathrm{Ba(OH)}_2\) reacts with 2 moles of \(\mathrm{HNO}_3\). Thus, 0.0146 moles of \(\mathrm{Ba(OH)}_2\) will react with: \[0.0146 \times 2 = 0.0292 \text{ moles of } \mathrm{HNO}_3.\]

Step 3: Calculate Volume of HNO_3 Solution Required

Given the concentration of the \(\mathrm{HNO}_3\) solution is 0.109 M. Use the formula \(\text{Concentration} = \frac{\text{Moles of Solute}}{\text{Volume of Solution (L)}}\) to find the required volume. Rearrange the formula to solve for volume: \[\text{Volume (L)} = \frac{0.0292 \text{ moles}}{0.109 \text{ M}} = 0.268 \text{ L.}\] Convert this to milliliters: \[0.268 \text{ L} \times 1000 \text{ mL/L} = 268 \text{ mL.}\]

Key Concepts

Molar Mass Calculation

Understanding molar mass is essential for stoichiometry calculations. Molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). To calculate the molar mass, sum the atomic masses of all the atoms in the chemical formula.
For example, the molar mass of \(\mathrm{Ba(OH)}_2\) involves combining the atomic masses of Barium (Ba), Oxygen (O), and Hydrogen (H). Barium has an atomic mass of 137.33 g/mol, Oxygen 16.00 g/mol, and Hydrogen 1.01 g/mol. Since there are two OH groups in \(\mathrm{Ba(OH)}_2\), you must account for both.
Thus, the calculation becomes:

• 137.33 for Ba
• + 2 \(\times\) (16.00 + 1.01) for the two OH groups
• = 171.35 g/mol.

With this molar mass, you can determine how many moles are in a given mass of \(\mathrm{Ba(OH)}_2\). For instance, with 2.50 grams of \(\mathrm{Ba(OH)}_2\), you calculate the moles as \(\frac{2.50}{171.35} = 0.0146\) moles.

Chemical Reactions

During a chemical reaction, substances known as reactants transform into different substances called products. In stoichiometry, understanding the proportion between these reactants and products is crucial. Each reactant and product in a chemical equation is represented by its molecular formula and the balanced coefficients reveal how many moles of each are needed or produced.
For example, the equation

• \(2 \mathrm{HNO}_3(\mathrm{aq}) + \mathrm{Ba(OH)}_2(\mathrm{s}) \rightarrow 2 \mathrm{H}_2\mathrm{O}(\ell) + \mathrm{Ba}\left(\mathrm{NO}_3\right)_2(\mathrm{aq})\)

shows that 2 moles of \(\mathrm{HNO}_3\) react with 1 mole of \(\mathrm{Ba(OH)}_2\) to produce 2 moles of \(\mathrm{H}_2\mathrm{O}\) and 1 mole of \(\mathrm{Ba(NO}_3)_2\).
Understanding these mole ratios helps solve problems about how much of one substance is needed to completely react with another.

Solution Concentration

Solution concentration, often expressed in molarity (M), indicates the number of moles of solute per liter of solution. Molarity is a vital concept when determining the volume of solution needed in a chemical reaction.
For instance, to find the volume of \(\mathrm{HNO}_3\) needed, with the molarity given as 0.109 M, use the formula:

• \(\text{Concentration (M)} = \frac{\text{Moles of Solute}}{\text{Volume of Solution (L)}}\)

Rearrange the formula to find the volume: \(\text{Volume (L)} = \frac{\text{Moles of Solute}}{\text{Molarity}}\).
So, for 0.0292 moles of \(\mathrm{HNO}_3\), the volume is \(\frac{0.0292}{0.109} = 0.268\) L, which converts to 268 mL.

Balanced Equation

Balancing equations is essential for understanding chemical reactions. It ensures that the number of atoms for each element is conserved in a reaction. This reflects the law of conservation of mass, stating that mass cannot be created or destroyed in an ordinary chemical reaction.
To balance an equation, adjust coefficients to ensure that atom counts for each element are equal on both sides. For example, in:

• \(2 \mathrm{HNO}_3(\mathrm{aq}) + \mathrm{Ba(OH)}_2(\mathrm{s}) \rightarrow 2 \mathrm{H}_2\mathrm{O}(\ell) + \mathrm{Ba}\left(\mathrm{NO}_3\right)_2(\mathrm{aq})\)

a balanced equation indicates 2 moles of \(\mathrm{HNO}_3\) will interact with 1 mole of \(\mathrm{Ba(OH)}_2\), producing 2 moles of water and 1 mole of \(\mathrm{Ba(NO}_3)_2\).
Balanced chemical equations provide a straightforward way to use stoichiometry to relate quantities of reactants and products.