Problem 89 - Chemistry The Central Science In Si Units

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 Mg(OH)_2 is 91.41%.

Step 1: Understand the Reaction

The reaction between magnesium hydroxide, \(\text{Mg(OH)}_2\), and hydrochloric acid, \(\text{HCl}\), is given by the balanced chemical equation: \[\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}.\] Reactions indicate that 1 mole of \(\text{Mg(OH)}_2\) reacts with 2 moles of \(\text{HCl}\).

Step 2: Calculate Initial Moles of HCl

Calculate the initial moles of \(\text{HCl}\) in the solution:\[\text{Moles of HCl} = 0.2050 \text{ M} \times 0.1000 \text{ L} = 0.0205 \text{ moles}.\]

Step 3: Determine Moles of Excess HCl

Calculate the moles of \(\text{NaOH}\) used for neutralization:\[\text{Moles of NaOH} = 0.1020 \text{ M} \times 0.01985 \text{ L} = 0.0020237 \text{ moles}.\]Since \(\text{NaOH}\) neutralizes \(\text{HCl}\) with a 1:1 ratio, the moles of excess \(\text{HCl}\) also equals 0.0020237 moles.

Step 4: Calculate Moles of HCl Reacted with Mg(OH)2

Subtract the excess \(\text{HCl}\) moles from the initial moles:\[\text{Moles of HCl that reacted} = 0.0205 \text{ moles} - 0.0020237 \text{ moles} = 0.0184763 \text{ moles}.\]

Step 5: Calculate Moles of Mg(OH)2

According to the reaction stoichiometry, 2 moles of \(\text{HCl}\) react with 1 mole of \(\text{Mg(OH)}_2\). Therefore:\[\text{Moles of Mg(OH)}_2 = \frac{0.0184763}{2} = 0.00923815 \text{ moles}.\]

Step 6: Calculate Mass of Mg(OH)2

Using the molar mass of \(\text{Mg(OH)}_2\) (58.3197 g/mol):\[\text{Mass of Mg(OH)}_2 = 0.00923815 \text{ moles} \times 58.3197 \text{ g/mol} = 0.539 \text{ g}.\]

Step 7: Calculate Percentage by Mass of Mg(OH)2

Finally, calculate the percentage by mass of \(\text{Mg(OH)}_2\) in the sample:\[\left(\frac{0.539 \text{ g}}{0.5895 \text{ g}} \right) \times 100\% = 91.41\%.\]

Key Concepts

Stoichiometry

Stoichiometry is the mathematical foundation that governs chemical reactions. It helps us understand the quantitative relationships between reactants and products. In this magnesium hydroxide analysis, stoichiometry is crucial.
We start by analyzing the balanced chemical reaction: \[\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}. \]This equation shows us the relationship: 1 mole of magnesium hydroxide reacts with 2 moles of hydrochloric acid (HCl).
This stoichiometric ratio allows us to solve the problem of determining how much Mg(OH)₂ is present by measuring the amount of HCl that reacts.

• Every mole of Mg(OH)₂ requires exactly 2 moles of HCl to fully react, which is crucial for determining how much of the magnesium hydroxide was present.
• The stoichiometric coefficients (the numbers in front of the molecules) in the balanced equation guide us when transforming moles of one chemical into moles of another.

By understanding stoichiometry, students can predict the outcomes of reactions and calculate required quantities of reactants or products.

Neutralization Reaction

Neutralization reactions are a type of chemical reaction where an acid reacts with a base to form water and a salt. In this case, the reaction involves hydrochloric acid \(\text{HCl}\) and magnesium hydroxide \(\text{Mg(OH)}_2\).
During the process, \[\text{Mg(OH)}_2\] neutralizes \[\text{HCl}\], indicating that it consumes the acidic properties by forming water \((\text{H}_2\text{O})\) and \[\text{MgCl}_2\].
This is a classic example of a neutralization reaction, where a base, \[\text{Mg(OH)}_2\], neutralizes the acid \[\text{HCl}\]:

• The reaction can yield a salt, which in this case is magnesium chloride \(\text{MgCl}_2\).
• This setup also involved subtracting out the amount of \[\text{HCl}\] that did not react with \[\text{Mg(OH)}_2\] to target only what actually participated in the reaction.
• This is done using remaining excess \[\text{HCl}\], which was later neutralized by \[\text{NaOH}\].

Hence, by measuring how much hydrochloric acid was neutralized, we indirectly determined how much magnesium hydroxide was present.

Molar Mass Calculation

Molar mass is a crucial concept that connects the microscopic world of atoms and molecules to the macroscopic world we can measure. For any compound, its molar mass can be calculated by adding the atomic masses of each element in its formula.
For magnesium hydroxide, \[\text{Mg(OH)}_2\], we calculate as follows:

• The atomic mass of Mg is approximately 24.31 g/mol.
• Each OH group has an oxygen (\(16.00 \, \text{g/mol}\)) and a hydrogen (\(1.01 \, \text{g/mol}\)), totaling \(17.01 \, \text{g/mol}\) each.
• With two OH groups, we sum these: \[2 \times 17.01 \, \text{g/mol} + 24.31 \, \text{g/mol} = 58.3197 \, \text{g/mol}.\]

Calculating the molar mass allows us to convert between moles and grams. Once we determine how many moles of \[\text{Mg(OH)}_2\] reacted, multiplying this by the molar mass gives us the mass in grams. This step is vital in finding out how much magnesium hydroxide is present in a sample, as observed through our calculations.