Question

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

Short answer

#tag_title#Step 2: Determine moles of reactants and the limiting reactant#tag_content# Moles of CaSiO₃ = $\frac{32.9\,\text{g}}{116.16\,\text{g/mol}} = 0.283\,\text{mol}$ Moles of HF = $\frac{31.8\,\text{L}\times 1.00\,\text{atm}}{0.08206\,\text{L}\cdot\text{atm/mol}\cdot\text{K}\times(27.0^{\circ}C+273.15)} = 1.32\,\text{mol}$ Limiting reactant: $\frac{1.32\,\text{mol}\,\text{HF}}{4} = 0.330\,\text{mol} \gt 0.283\,\text{mol}\,\text{CaSiO}_{3}$, so CaSiO₃ is the limiting reactant. #tag_title#Step 3: Calculate mass of products formed#tag_content# Moles of SiF₄ = moles of CaSiO₃ = 0.283 mol Moles of H₂O = $2\times0.283\,\text{mol} = 0.566\,\text{mol}$ Mass of SiF₄ = $0.283\,\text{mol} \times 104.08\,\text{g/mol} = 29.4\,\text{g}$ Mass of H₂O = $0.566\,\text{mol} \times 18.015\,\text{g/mol} = 10.2\,\text{g}$ So, 29.4 g of SiF₄ and 10.2 g of H₂O are produced in the reaction.

Step-by-step solution

Balance the chemical equation

Balancing the given equation: $$CaSiO_3(s)+4HF(g) \longrightarrow CaF_2(aq)+SiF_4(g)+2H_2O(l)$$

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial step in the study of chemical reactions. It ensures that matter is neither created nor destroyed according to the Law of Conservation of Mass. For a chemical equation to be balanced, the number of atoms for each element on the reactant side must be equal to the number of atoms on the product side.

Let's take a look at the unbalanced equation given in the exercise: \[ \mathrm{CaSiO}_{3}(s)+\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(aq)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2}\mathrm{O}(l) \]To balance this, we adjust the coefficients of the reactants and products until the atoms are equal on both sides:

• Start by balancing the elements that appear in only one reactant and one product. For example, balance calcium (Ca) and silicon (Si) first since they appear in one part on both sides.
• Once the easy elements are balanced, adjust the hydrogen and oxygen by ensuring H and O are equal in quantity on both sides.

By performing this process, it becomes: \[ \mathrm{CaSiO}_{3}(s)+4\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(aq)+\mathrm{SiF}_{4}(g)+2\mathrm{H}_{2}\mathrm{O}(l) \] This balanced equation confirms that you have the same number of each type of atom on either side of the equation.

Stoichiometry

Stoichiometry involves using the balanced equation to find the relationships between the reactants and products in a chemical reaction. It tells us how much of each compound we need or can produce.

From the balanced equation, \[ \mathrm{CaSiO}_{3}(s)+4\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(aq)+\mathrm{SiF}_{4}(g)+2\mathrm{H}_{2}\mathrm{O}(l) \] we can derive the molar ratios between different substances. For instance:

• 1 mole of \(\mathrm{CaSiO}_3\) reacts with 4 moles of \(\mathrm{HF}\).
• This produces 1 mole of \(\mathrm{CaF_2}\), 1 mole of \(\mathrm{SiF_4}\), and 2 moles of \(\mathrm{H_2O}\).

These ratios are crucial for calculating how much product forms when you begin with a given amount of reactant, thereby allowing the determination of yields in real-world reactions.

Limiting Reactant

In a chemical reaction, the limiting reactant is the one that is consumed first, thus determining the maximum amount of product that can be formed. It's pivotal to identify this because it allows chemists to calculate the theoretical yield of the reaction.

To identify the limiting reactant between \(\mathrm{CaSiO}_3\) and \(\mathrm{HF}\) in the exercise:

• Calculate the moles of each reactant. Using the given grams or liters, convert these values into moles using their molar mass or ideal gas law for gases.
• Compare the mole ratio of each reactant to the balanced equation. The reactant that provides the lesser amount of product is the limiting reactant.

The reaction can go no further once the limiting reactant is used up completely.

Ideal Gas Law

The Ideal Gas Law is a useful tool when dealing with gases. It relates the pressure, volume, temperature, and amount of a gas through the equation:\[ PV = nRT \]where:

• \(P\) is the pressure in atmospheres.
• \(V\) is the volume in liters.
• \(n\) is the number of moles.
• \(R\) is the ideal gas constant (0.0821 L.atm/mol.K).
• \(T\) is the temperature in Kelvin.

To calculate the number of moles of \(\mathrm{HF}\), you can rearrange to:\[ n = \frac{PV}{RT} \]This formula helps determine the starting moles of gas reactants and thereby aids in identifying the limiting reactant. It's especially helpful when dealing with standard conditions, as shown in this exercise.