Problem 148 - Chemistry An Atoms First Approach

Question

Consider the unbalanced chemical equation below: $$\mathrm{CaSiO}_{3}(s)+\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Suppose a \(32.9-\mathrm{g}\) sample of \(\mathrm{CaSiO}_{3}\) is reacted with \(31.8 \mathrm{L}\) of \(\mathrm{FH}\) at \(27.0^{\circ} \mathrm{C}\) and \(1.00\) atm. Assuming the reaction goes to completion, calculate the mass of the \(\mathrm{SiF}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced in the reaction.

Step-by-Step Solution

Answer

In conclusion, the mass of SiF4 and H2O produced in the reaction is 34.178 g and 5.915 g, respectively.

Step 1: Balance the chemical equation

We have the unbalanced chemical equation: \( \mathrm{CaSiO}_{3}(s) + \mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(aq) + \mathrm{SiF}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(l) \) To balance this equation, we need equal numbers of each atom on both sides. This can be achieved by multiplying the reactants and products with appropriate coefficients: \( \mathrm{CaSiO}_{3}(s) + 4\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(aq) + \mathrm{SiF}_{4}(g) + 2\mathrm{H}_{2} \mathrm{O}(l) \) Now the equation is balanced.

Step 2: Find the moles of each reactant

We can use the given mass of CaSiO_3 and the number of moles from the Ideal Gas Law to find the moles of each reactant. For CaSiO_3, we have mass = 32.9 g and molecular weight = (40.08 + 28.09 + 3 × 16.00) g/mol = 100.17 g/mol, and thus we have: Moles of CaSiO_3 = (mass) / (molecular weight) = 32.9 g / 100.17 g/mol = 0.3285 mol For HF, we have volume = 31.8 L, pressure = 1.00 atm, and temperature =27.0 °C, and we can use the Ideal Gas Law: \( PV = nRT \), where R = 0.0821 L atm / mol K First, convert the temperature from Celsius to Kelvin: T = 27.0 °C + 273.15 = 300.15 K Now we can find the moles of HF: Moles of HF = (PV) / (RT) = (1.00 atm × 31.8 L) / (0.0821 L atm / mol K × 300.15 K) = 1.3032 mol

Step 3: Find the limiting reactant

Now we need to find the limiting reactant; it will control how much of the products can be produced. To do this, compare the mole ratios of the reactants: Mole ratio of CaSiO_3 to HF: 1:4 Actual mole ratio: 0.3285 mol CaSiO_3 / 1.3032 mol HF = 0.2519 Since the actual mole ratio is less than the required mole ratio, CaSiO_3 is the limiting reactant.

Step 4: Calculate the mass of SiF4 and H2O produced

To find the mass of products, we need to use stoichiometry based on the balanced chemical equation and the limiting reactant: Mass of SiF4 produced = (Moles of CaSiO_3 × mol SiF4 / mol CaSiO_3 ) × Molecular weight of SiF4 = (0.3285 mol × 1 mol SiF4 / 1 mol CaSiO_3) × (28.09 + 4 × 19.00) g/mol = 0.3285 mol × 104.09 g/mol = 34.178 g Mass of H2O produced = (Moles of CaSiO_3 × mol H2O / mol CaSiO_3) × Molecular weight of H2O = (0.3285 mol × 2 mol H2O / 1 mol CaSiO_3) × (2 × 1.01+ 16.00) g/mol = 0.3285 mol × 18.02 g/mol = 5.915 g In conclusion, the mass of SiF4 and H2O produced in the reaction is 34.178 g and 5.915 g, respectively.

Key Concepts

Understanding Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction. By displaying the reactants and products, it shows how substances transform during a reaction. The beauty of chemical equations lies in their ability to succinctly convey complex reactions using simple formulas.

To comprehend them better, here are a few key points:

• Reactants are the starting chemicals depicted on the left side of the equation.
• Products are the substances formed, displayed on the right side.
• Coefficients before each chemical formula indicate the number of molecules involved.

Balancing chemical equations is crucial because it ensures the law of conservation of mass, meaning the number of each type of atom must be the same on both sides of the equation. This reflects that matter cannot be created or destroyed in a chemical reaction.

For instance, in our exercise, the equation is initially unbalanced. Balancing it involves adding coefficients where necessary, transforming it from:
\( ext{CaSiO}_3(s) + ext{HF}(g) ightarrow ext{CaF}_2(aq) + ext{SiF}_4(g) + ext{H}_2 ext{O}(l) \) to
\( ext{CaSiO}_3(s) + 4 ext{HF}(g) ightarrow ext{CaF}_2(aq) + ext{SiF}_4(g) + 2 ext{H}_2 ext{O}(l) \).

This balancing provides clarity on the exact quantities of reactants needed and products formed.

Limiting Reactant Deep Dive

In chemical reactions, the limiting reactant is the substance that gets completely used up first, halting further reaction because no more product can be formed. This concept is vital for predicting the maximum amount of product that can be generated.

Here’s how to pinpoint the limiting reactant:

• Use the balanced chemical equation to find the mole ratio of reactants.
• Convert the amounts of each reactant to moles.
• Compare these moles to the required mole ratios from the equation.

In our exercise, we found 0.3285 moles of \( ext{CaSiO}_3 \) and 1.3032 moles of \( ext{HF} \). With a required ratio of 1:4, \( ext{CaSiO}_3 \) actually dictates the amount of product formed because its actual mole ratio is lower than necessary.

This means that while \( ext{HF} \) is present in excess, it cannot be fully utilized due to the limited quantity of \( ext{CaSiO}_3 \). Recognizing the limiting reactant is like knowing when to stop pouring vinegar into baking soda in a fizz experiment – once either is depleted, the reaction ceases.

Exploring the Ideal Gas Law

The Ideal Gas Law is a cornerstone of chemistry, providing a relationship between pressure, volume, temperature, and the amount of gas. The equation is expressed as:
\( PV = nRT \)

Where:

• \( P \) stands for pressure measured in atmospheres (atm).
• \( V \) represents volume measured in liters (L).
• \( n \) is the amount of substance in moles.
• \( R \) is the ideal gas constant, 0.0821 L atm/mol K.
• \( T \) is the temperature in Kelvin (K).

To solve problems involving gases, proper conversions are key. For instance, temperature in Celsius should be converted to Kelvin by adding 273.15. The gas law helps determine the number of moles, crucial for stoichiometric calculations in reactions involving gases.

In the step-by-step solution of our exercise, we plug in the values: \( P = 1.00 \text{ atm} \), \( V = 31.8 \text{ L} \), and \( T = 300.15 \text{ K} \), leading to the calculation of 1.3032 moles of \( ext{HF} \).

The Ideal Gas Law thus acts as a bridge between the macroscopic world (what we measure) and the microscopic view (moles of gas), allowing predictions about how gases will behave under different conditions.