Problem 89 - Chemistry The Central Science

Question

A \(0.5895-\) g sample of impure magnesium hydroxide is dissolved in 100.0 \(\mathrm{mL}\) of 0.2050 \(\mathrm{M} \mathrm{HCl}\) solution. The excess acid then needs 19.85 \(\mathrm{mL}\) of 0.1020 \(\mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percentage by mass of magnesium hydroxide in the sample, assuming that it is the only substance reacting with the HCl solution.

Step-by-Step Solution

Answer

The percentage by mass of magnesium hydroxide in the sample is approximately 91.34%.

Step 1: Write down the balanced chemical equations

We have two reactions taking place here: 1. The reaction between magnesium hydroxide (Mg(OH)\(_2\)) and hydrochloric acid (HCl): \[ Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O \] 2. The reaction between sodium hydroxide (NaOH) and the excess hydrochloric acid (HCl): \[ NaOH + HCl \rightarrow NaCl + H_2O \]

Step 2: Calculate the moles of HCl initially present

We are given the volume (100.0 mL) and concentration (0.2050 M) of the HCl solution. To find the moles of HCl, we can use the formula: Moles of HCl = volume × concentration Moles of HCl = \( (100.0 \times 10^{-3} \, \text{L}) (0.2050 \, \text{M}) \) Moles of HCl = \( 0.02050 \, \text{mol} \)

Step 3: Calculate the moles of HCl that reacted with Mg(OH)\(_2\)

We know that there are 19.85 mL of 0.1020 M NaOH needed to neutralize the excess HCl. We can find the moles of HCl reacted with NaOH as follows: Moles of HCl reacted with NaOH = volume × concentration Moles of HCl reacted with NaOH = \( (19.85 \times 10^{-3} \, \text{L}) (0.1020 \, \text{M}) \) Moles of HCl reacted with NaOH = \( 0.002023 \, \text{mol} \) Now, subtract this amount of HCl from the initial moles of HCl to find the moles of HCl that reacted with Mg(OH)\(_2\): Moles of HCl reacted with Mg(OH)\(_2\) = Initial moles of HCl - Moles of HCl reacted with NaOH Moles of HCl reacted with Mg(OH)\(_2\) = 0.02050 - 0.002023 Moles of HCl reacted with Mg(OH)\(_2\) = \( 0.01848 \, \text{mol} \)

Step 4: Calculate the moles of Mg(OH)\(_2\) in the sample

Using the stoichiometry of the balanced equation of reaction between Mg(OH)\(_2\) and HCl (1:2 ratio), we can find the moles of Mg(OH)\(_2\): Moles of Mg(OH)\(_2\) = Moles of HCl reacted with Mg(OH)\(_2\) / 2 Moles of Mg(OH)\(_2\) = 0.01848 / 2 Moles of Mg(OH)\(_2\) = \( 0.00924 \, \text{mol} \)

Step 5: Calculate the mass of Mg(OH)\(_2\) in the sample

To find the mass of Mg(OH)\(_2\), we multiply its moles by its molar mass: Mass of Mg(OH)\(_2\) = moles × molar mass Molar mass of Mg(OH)\(_2\) = 24.31 (Mg) + 2(16.00 + 1.01) (2OH) = 58.33 g/mol Mass of Mg(OH)\(_2\) = \( (0.00924 \, \text{mol}) (58.33 \, \text{g/mol}) \) Mass of Mg(OH)\(_2\) = \( 0.5385 \, \text{g} \)

Step 6: Calculate the percentage by mass of Mg(OH)\(_2\) in the sample

Finally, we can find the percentage by mass of Mg(OH)\(_2\) in the sample using the formula: Percentage by mass of Mg(OH)\(_2\) = (Mass of Mg(OH)\(_2\) / Total mass of the sample) × 100 Percentage by mass of Mg(OH)\(_2\) = \( (0.5385 / 0.5895) \times 100 \) Percentage by mass of Mg(OH)\(_2\) = \( 91.34 \% \) Thus, the percentage by mass of magnesium hydroxide in the sample is approximately 91.34%.

Key Concepts

Chemical Reactions

In this exercise, two key chemical reactions take place involving hydrochloric acid (HCl). The first reaction involves magnesium hydroxide (Mg(OH)\(_2\)), which reacts with HCl to form magnesium chloride (MgCl\(_2\)) and water. The balanced chemical equation is: \[ Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O \] Understanding the stoichiometry of this reaction is crucial. Here, each molecule of Mg(OH)\(_2\) requires two molecules of HCl to react completely.
The second reaction involves the excess HCl reacting with sodium hydroxide (NaOH). The reaction forms sodium chloride (NaCl) and water: \[ NaOH + HCl \rightarrow NaCl + H_2O \] This reaction is simpler, with a one-to-one ratio. It’s important to track these reactions to understand which reactants are consumed or left in excess.

Molar Mass

Molar mass is a critical concept for calculating the mass of compounds in stoichiometry problems. It is the mass of one mole of a substance, expressed in grams per mole (g/mol).
For magnesium hydroxide, the molar mass is calculated by adding the atomic masses of its constituent atoms:

• Magnesium (Mg): 24.31 g/mol
• Oxygen (O): 16.00 g/mol (2 atoms)
• Hydrogen (H): 1.01 g/mol (2 atoms)

The molar mass of Mg(OH)\(_2\) becomes: \[ 24.31 + 2(16.00 + 1.01) = 58.33 \, \text{g/mol} \] Knowing this value helps convert between moles and grams, like finding the mass of Mg(OH)\(_2\) that reacted in the problem.

Neutralization

Neutralization is a vital process where an acid reacts with a base to produce water and a salt. In this exercise, two neutralization reactions occur.
The primary reaction involves Mg(OH)\(_2\) neutralizing HCl. Magnesium hydroxide, a base, reacts with hydrochloric acid to produce water and magnesium chloride.
After all the Mg(OH)\(_2\) reacts, some HCl remains. It is then neutralized by NaOH, another base. This second neutralization helps determine the excess HCl leftover. Neutralization is used to calculate how much of the acid reacted, which indirectly tells us the amount of base (Mg(OH)\(_2\)) in the sample.

Percentage Composition

Percentage composition indicates how much of a specific component is present in a compound or mixture. In this problem, it is used to find the mass percentage of magnesium hydroxide in the impure sample.
The formula for percentage composition is: \[ \text{Percentage by mass} = \left( \frac{\text{Mass of component}}{\text{Total mass of sample}} \right) \times 100 \] Knowing the mass of Mg(OH)\(_2\), calculated as 0.5385 g, and the total sample mass of 0.5895 g, we plug into the formula to find: \[ \text{Percentage by mass of Mg(OH)}_2 = \left( \frac{0.5385}{0.5895} \right) \times 100 \approx 91.34\% \] This tells us that approximately 91.34% of the sample is pure magnesium hydroxide, giving insight into its purity.