Problem 61 - Chemistry The Central Science In Si Units

Question

Hydrofluoric acid, HF \((a q)\), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the \(\mathrm{HF}(a q)\). Sodium silicate \(\left(\mathrm{Na}_{2} \mathrm{SiO}_{3}\right)\), for example, reacts as follows: $$ \mathrm{Na}_{2} \mathrm{SiO}_{3}(s)+8 \mathrm{HF}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SiF}_{6}(a q)+2 \mathrm{NaF}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) How many moles of HF are needed to react with 0.300 mol of \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (b) How many grams of NaF form when \(0.500 \mathrm{~mol}\) of HF reacts with excess \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (c) How many grams of \(\mathrm{Na}_{2} \mathrm{SiO}_{3}\) can react with \(0.800 \mathrm{~g}\) of HF?

Step-by-Step Solution

Answer

(a) 2.4 moles HF, (b) 5.25 g NaF, (c) 0.6103 g Na2SiO3.

Step 1: Determine Moles of HF for Reaction

From the chemical equation, 1 mole of \( \mathrm{Na}_2 \mathrm{SiO}_3 \) reacts with 8 moles of \( \mathrm{HF} \). For 0.300 moles of \( \mathrm{Na}_2 \mathrm{SiO}_3 \), the moles of \( \mathrm{HF} \) needed are \( 0.300 \, \text{mol} \times 8 = 2.4 \, \text{mol} \) HF.

Step 2: Convert Moles HF to Grams of NaF

According to the reaction, 8 moles of \( \mathrm{HF} \) produce 2 moles of \( \mathrm{NaF} \). The molar ratio of \( \mathrm{HF} \) to \( \mathrm{NaF} \) is 4:1. Thus, 0.500 mol of \( \mathrm{HF} \) produces \( \frac{0.500}{4} = 0.125 \, \text{mol} \) \( \mathrm{NaF} \). The molar mass of \( \mathrm{NaF} \) is 42.00 g/mol. Therefore, the mass is \( 0.125 \, \text{mol} \times 42.00 \, \text{g/mol} = 5.25 \, \text{g} \) \( \mathrm{NaF} \).

Step 3: Determine Moles of HF for Reaction with NaSiO3

To find how many grams of \( \mathrm{Na}_2 \mathrm{SiO}_3 \) can react with 0.800 g of HF, we first convert grams to moles of HF: \( \frac{0.800 \ ext{g}}{20.01 \ ext{g/mol}} = 0.040 \, \text{mol HF} \). From the mole ratio \( 8 \text{HF} : 1 \text{Na}_2 \text{SiO}_3 \), the moles of \( \mathrm{Na}_2 \mathrm{SiO}_3 \) required is \( \frac{0.040}{8} = 0.005 \, \text{mol} \).

Step 4: Convert Moles of NaSiO3 to Grams

The molar mass of \( \mathrm{Na}_2 \mathrm{SiO}_3 \) is approximately 122.06 g/mol. Thus, 0.005 mol of \( \mathrm{Na}_2 \mathrm{SiO}_3 \) is \( 0.005 \, \text{mol} \times 122.06 \, \text{g/mol} = 0.6103 \, \text{g} \).

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. By using stoichiometry, we can calculate how much of each substance is needed or produced in a given reaction.
In the original exercise, we are working with the reaction between Hydrofluoric acid (HF) and Sodium silicate (Na\(_2\)SiO\(_3\)). According to the balanced chemical equation:
\[ \text{Na}_2\text{SiO}_3(s) + 8\text{HF}(aq) \rightarrow \text{H}_2\text{SiF}_6(aq) + 2\text{NaF}(aq) + 3\text{H}_2\text{O}(l) \]
Here, stoichiometry shows us that 1 mole of Na\(_2\)SiO\(_3\) reacts with 8 moles of HF.

• This means that if you have a certain amount of Na\(_2\)SiO\(_3\) and want to know how much HF is required, you would multiply the moles of Na\(_2\)SiO\(_3\) by 8.
• For example, to react with 0.300 moles of Na\(_2\)SiO\(_3\), you need 0.300 mol \( \times 8 = 2.4 \) moles of HF.

Understanding stoichiometry is crucial because it allows chemists to predict the outcomes of reactions—how much product will form or how much reactant will be needed.

Molar Mass

Molar mass is an important concept that refers to the mass of one mole of a given substance. It is expressed in grams per mole (g/mol), and it is crucial for converting between moles and grams, which is often necessary in chemical calculations.
For example, in the reaction given:
To find the mass of NaF produced, you first calculate the moles and then use the molar mass of NaF. The formula for NaF is made up of one sodium atom (Na) and one fluorine atom (F), so:

• The molar mass of Na is approximately 22.99 g/mol.
• The molar mass of F is approximately 19.00 g/mol.
• Therefore, the molar mass of NaF is 22.99 g/mol + 19.00 g/mol = 42.00 g/mol.

With this information, we convert the moles of NaF to grams using the molar mass. For example, 0.125 moles of NaF would weigh 0.125 mol \( \times 42.00 \text{ g/mol} = 5.25 \text{ g}. \)
Recognizing molar mass allows for efficient conversion between moles and grams, making it possible to understand and predict the reaction results in practical terms.

Conversion of Units

Conversion of units is a common practice in chemistry that involves changing one unit of measure to another while keeping the value equal. This is particularly important in chemical calculations where values are often provided in different units.
In our exercise, to solve for the mass of Na\(_2\)SiO\(_3\) that can react with a given mass of HF, a conversion includes:

• Starting with the mass of HF given in grams (e.g., 0.800 g of HF).
• Using the molar mass of HF, which is approximately 20.01 g/mol, to convert the mass in grams to moles: \( 0.800 \text{ g} \div 20.01 \text{ g/mol} = 0.040 \text{ mol HF}. \)

This conversion is crucial for applying stoichiometric ratios, as calculations generally require working with them in moles.
Following this, we apply the mole ratios to determine how much of another reactant, such as Na\(_2\)SiO\(_3\), can react with the given amount of HF. Always remember the connections between grams, moles, and volumes, especially when dealing with different states of matter.

Acid-Base Chemistry

Acid-base chemistry involves understanding the nature and behavior of acids and bases, along with how they react with each other. Hydrofluoric acid \((\text{HF})\) is an example of a weak acid often found attacking substances due to its high reactivity.
In the chemical equation given, HF is reacting with sodium silicate (Na\(_2\)SiO\(_3\)) and is part of a broader category of reactions called acid-base reactions. The role of HF in this context involves its acidic properties, which allows it to interact aggressively with the silicate compounds found in glass.
Key considerations in acid-base chemistry include:

• The strength of acids and bases, which are measured by their ability to donate or accept protons (H\(^+\)).
• Reactions of acids with bases typically result in the formation of a salt and water.

In this reaction:

• HF acts on Na\(_2\)SiO\(_3\) to form H\(_2\)SiF\(_6\), sodium fluoride (NaF), and water \((\text{H}_2\text{O})\), demonstrating a classical transformation seen in acid-base interactions.

Understanding acid-base chemistry helps us predict and control how different substances interact, which is vital in both industrial and laboratory settings.