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 \(\mathrm{Ba}(\mathrm{OH})_{2} ?\) $$2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{s}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})$$

Step-by-Step Solution

Answer

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

Step 1: Convert Mass to Moles

First, calculate the number of moles of \( \mathrm{Ba(OH)}_2 \) using its molar mass. The molar mass of \( \mathrm{Ba(OH)}_2 \) is \( 137.33 + 2(16.00 + 1.01) = 171.35 \text{ g/mol} \). The moles of \( \mathrm{Ba(OH)}_2 \) are calculated as \( \frac{2.50 \text{ g}}{171.35 \text{ g/mol}} \approx 0.0146 \text{ mol} \).

Step 2: Use the Stoichiometry of the Reaction

From the balanced chemical equation, \( 1 \text{ mole of } \mathrm{Ba(OH)}_2 \) reacts with \( 2 \text{ moles of } \mathrm{HNO}_3 \). Thus, the moles of \( \mathrm{HNO}_3 \) required are \( 0.0146 \text{ mol} \times 2 = 0.0292 \text{ mol} \).

Step 3: Calculate the Volume of \( \mathrm{HNO}_3 \) Solution

Use the molarity equation \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \) to find the volume required. Rearrange to find volume: \( \text{Volume} = \frac{\text{moles}}{M} \). Substituting the values, \( \text{Volume} = \frac{0.0292 \text{ mol}}{0.109 \text{ M}} \approx 0.2673 \text{ L} \). Convert this volume into milliliters by multiplying by 1000, resulting in \( 267.3 \text{ mL} \).

Key Concepts

Molarity

Molarity is a way of expressing the concentration of a solution. It is represented by the letter "M" and is calculated using the formula: \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). Molarity tells us how many moles of a substance, or solute, are present in one liter of solution.
When you know the molarity, you can determine how much of a solution is needed to react with a certain amount of another substance. In the exercise, we had a solution of nitric acid \((\mathrm{HNO}_3)\) with a molarity of 0.109 M. Using this information helps us find the volume required to react with a different chemical. It's like using a map to find the shortest way to where you want to go in chemistry terms.

• Makes measuring chemical reactions easy.
• Allows precise calculations in experiments.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into different substances called products. The key part of a chemical reaction is the way substances interact and change to form new substances.
In our exercise, the chemical reaction between \(\mathrm{HNO}_3\) and \(\mathrm{Ba(OH)}_2\) produces water \((\mathrm{H}_2\mathrm{O})\) and barium nitrate \((\mathrm{Ba(NO}_3)_2)\). Each reaction has its own unique equation that shows how materials change. Understanding chemical reactions involves knowing not just what substances will be produced, but also the exact amounts needed.

• Involves breaking and forming bonds between atoms.
• Can be affected by conditions like temperature and pressure.

Balanced Chemical Equation

A balanced chemical equation is a vital concept in chemistry. It ensures that the number of atoms for each element is the same on both sides of the equation. This reflects the law of conservation of mass, meaning mass cannot be created or destroyed in a reaction.
In the balanced equation for our exercise, \(2 \mathrm{HNO}_3(\mathrm{aq}) + \mathrm{Ba(OH)}_2(\mathrm{s}) \rightarrow 2 \mathrm{H}_2\mathrm{O}(\ell) + \mathrm{Ba}(\mathrm{NO}_3)_2(\mathrm{aq})\), the number of nitrogen, oxygen, hydrogen, and barium atoms are equal on both sides. Without a balanced equation, it would be impossible to know the precise amounts of reactants needed or products formed. It’s like having a detailed recipe—without it, the meal (or reaction) could be a disaster!

• Ensures everyone is "following the rules" of chemistry.
• Essential for calculating reactant and product quantities.