Problem 109 - Chemistry The Central Science

Question

A sample of \(5.53 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is added to \(25.0 \mathrm{~mL}\) of 0.200 \(\mathrm{M} \mathrm{HNO}_{3}\) (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) How many moles of \(\mathrm{Mg}(\mathrm{OH})_{2}, \mathrm{HNO}_{3},\) and \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) are present after the reaction is complete?

Step-by-Step Solution

Answer

The balanced chemical equation for the reaction between magnesium hydroxide and nitric acid is \( Mg(OH)_2 + 2 HNO_3 \rightarrow Mg(NO_3)_2 + 2 H_2O \). HNO₃ is the limiting reactant. After the reaction is complete, there are approximately 0.0923 moles of Mg(OH)₂, 0 moles of HNO₃, and 0.00250 moles of Mg(NO₃)₂.

Step 1: 1. Write the balanced chemical equation

The balanced chemical equation for the neutralization reaction between magnesium hydroxide and nitric acid is given by: \( Mg(OH)_2 + 2 HNO_3 \rightarrow Mg(NO_3)_2 + 2 H_2O \) This equation indicates that for every mole of magnesium hydroxide reacted, two moles of nitric acid are also reacted, forming one mole of magnesium nitrate and two moles of water.

Step 2: 2. Calculate the moles of Mg(OH)₂ and HNO₃

First, we need to calculate the moles of magnesium hydroxide and nitric acid present in the reaction. Moles of Mg(OH)₂ = mass / molar mass The molar mass of Mg(OH)₂ = 24.31 (Mg) + 2 * (16.00 (O) + 1.01 (H)) = 58.33 g/mol Moles of Mg(OH)₂ = 5.53 g / 58.33 g/mol ≈ 0.0948 mol Moles of HNO₃ = concentration * volume To convert the volume from milliliters to liters: 25.0 mL * (1L/1000mL) = 0.025 L Moles of HNO₃ = (0.200 mol/L) * 0.025 L = 0.00500 mol

Step 3: 3. Identify the limiting reactant

To find out which reactant is the limiting reactant, we compare the moles ratios from the balanced chemical equation to the initial moles we have: Mg(OH)₂ : HNO₃ required for reaction = 1 : 2 Now we must compare the ratio of initial moles of Mg(OH)₂ and HNO₃ to this mol ratio from the balanced equation: 0.0948 mol Mg(OH)₂ : 0.00500 mol HNO₃ We can see that there are not enough moles of HNO₃ relative to Mg(OH)₂. Thus, HNO₃ is the limiting reactant.

Step 4: 4. Calculate the moles of Mg(OH)₂, HNO₃, and Mg(NO₃)₂ after the reaction

As HNO₃ is the limiting reactant, we use its moles to determine the amounts of the other compounds: (a) Mg(OH)₂ From the stoichiometry, 1 mole of Mg(OH)₂ reacts with 2 moles of HNO₃. Therefore, Moles of Mg(OH)₂ reacted = Moles of HNO₃ / 2 = 0.00500 mol / 2 = 0.00250 mol Moles of Mg(OH)₂ remaining = initial moles - reacted moles = 0.0948 mol - 0.00250 mol ≈ 0.0923 mol (b) HNO₃ Since HNO₃ is the limiting reactant, all of its moles are consumed by the reaction. Moles of HNO₃ remaining = 0 mol (c) Mg(NO₃)₂ From the stoichiometry, 1 mole of Mg(OH)₂ reacts with 2 moles of HNO₃ to form 1 mole of Mg(NO₃)₂. Thus, the moles of Mg(NO₃)₂ formed are equal to the moles of Mg(OH)₂ reacted. Moles of Mg(NO₃)₂ formed = 0.00250 mol In summary, after the reaction is complete, there are approximately 0.0923 moles of Mg(OH)₂, 0 moles of HNO₃, and 0.00250 moles of Mg(NO₃)₂.

Key Concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely consumed first. It determines the amount of product that can be formed. In our reaction between magnesium hydroxide, \( \text{Mg(OH)}_2 \), and nitric acid, \( \text{HNO}_3 \), identifying the limiting reactant is crucial for accurate stoichiometric calculations.
When the moles of available reactants are compared to the stoichiometric ratio from the balanced equation, we see which reactant will run out first. Hence, it acts as a bottleneck for the reaction.

• Consider the balanced equation: \( \text{Mg(OH)}_2 + 2 \text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + 2 \text{H}_2\text{O} \).
• The balanced ratio is 1 mole of Mg(OH)₂ to 2 moles of HNO₃.
• Here, \(0.0948\) mol of \(\text{Mg(OH)}_2\) and \(0.00500\) mol of \(\text{HNO}_3\) are present.
• \( \text{HNO}_3 \) is the limiting reactant because it will run out before \( \text{Mg(OH)}_2 \) does, based on the stoichiometry given by the balanced equation.

When performing a reaction, always start by identifying the limiting reactant to predict the extent of the reaction accurately.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the conservation of mass and the mole concept.In the given reaction between \( \text{Mg(OH)}_2 \) and \( \text{HNO}_3 \), stoichiometry allows us to predict the quantities of products based on the reactants. Start by ensuring you have the balanced chemical equation as your roadmap. It provides molar ratios required for calculations, ensuring all calculations adhere to the law of conservation of mass.
For step-by-step stoichiometry calculations:

• Start with known quantities: for example, grams of \( \text{Mg(OH)}_2 \) or volume and concentration of \( \text{HNO}_3 \).
• Convert these known quantities into moles using molar mass and molarity, respectively.
• Employ the balanced chemical equation to convert moles of reactants to moles of desired products or remaining reactants.
• Every calculation should reflect the stoichiometric coefficients from the equation.

Understanding stoichiometry is crucial to tackle questions about how much of each substance is present before and after the reaction. Accurate stoichiometric calculations ensure that we fully comprehend the flow and interaction of matter in chemical processes.

Balanced Chemical Equation

A balanced chemical equation is an expression representing a chemical reaction that shows the same number of each type of atom on both sides of the equation. This reflects the law of conservation of mass, where no atoms are lost or gained in a reaction.
In our case, balancing the equation \( \text{Mg(OH)}_2 + 2 \text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + 2 \text{H}_2\text{O} \) determines the precise stoichiometric relationships. Each side of the reaction has the same types and numbers of atoms, ensuring precise calculations in stoichiometry.
Why balancing is key:

• Ensures the reaction adheres to the conservation of mass.
• Provides the correct stoichiometric ratios needed for calculating product quantities.
• Helps identify the limiting and excess reactants, guiding reaction predictions.

Balancing chemical equations is foundational in understanding chemical reactions. It is a necessary skill for further calculations and interpreting chemical changes effectively. Proper balancing leads to accurate and predictable outcomes in both theoretical and practical chemistry.