Problem 131 - Chemistry: The Central Science in SI Units, Global Edition

Question

One method to synthesize ionic solids is by the heating of two reactants at high temperatures. Consider the reaction of \(\mathrm{MgO}\) with \(\mathrm{TiO}_{2}\) to form \(\mathrm{MgTiO}_{3}\). Determine the amount of each of the two reactants to prepare \(3.250 \mathrm{~g} \mathrm{MgTiO}_{3}\), assuming the reaction goes to completion. (a) Write a balanced chemical reaction. (b) Calculate the formula weight of \(\mathrm{MgTiO}_{3}\). (c) Determine the moles of \(\mathrm{MgTiO}_{3}\). (d) Determine moles and mass (g) of MgO required. (e) Determine moles and mass (g) of \(\mathrm{TiO}_{2}\) required.

Step-by-Step Solution

Answer

The balanced chemical reaction is MgO + TiO₂ → MgTiO₃. The formula weight of MgTiO₃ is 120.18 g/mol. To prepare 3.250 g of MgTiO₃, we need 0.02703 mol of MgTiO₃. Therefore, we will require 1.090 g of MgO and 2.160 g of TiO₂ for the reaction.

Step 1: (a) Write a balanced chemical reaction

The equation of the reaction between magnesium oxide (MgO) and titanium dioxide (TiO₂) to produce magnesium titanate (MgTiO₃) is given below: MgO + TiO₂ → MgTiO₃ The given reaction equation is already balanced.

Step 2: (b) Calculate the formula weight of MgTiO₃

To find the formula weight of magnesium titanate (MgTiO₃), we sum up the atomic weights of its individual elements: Formula weight of MgTiO₃ = Atomic weight of Mg + Atomic weight of Ti + (3 × Atomic weight of O) Atomic weights: Mg(24.31 g/mol), Ti(47.87 g/mol) and O(16.00 g/mol) Formula weight of MgTiO₃ = 24.31 + 47.87 + (3 × 16.00) = 24.31 + 47.87 + 48 = 120.18 g/mol

Step 3: (c) Determine the moles of MgTiO₃

The mass of MgTiO₃ is given as 3.250 g. To find the moles of MgTiO₃, we will use the formula: Moles = Mass / Formula weight Moles of MgTiO₃ = \(\frac{3.250 \thinspace g}{120.18 \thinspace g/mol}\) = 0.02703 mol

Step 4: (d) Determine moles and mass (g) of MgO required

From the balanced chemical reaction, 1 mole of MgO reacts with 1 mole of TiO₂ to form 1 mole of MgTiO₃. So to form 0.02703 mol of MgTiO₃, 0.02703 mol of MgO is required. To find the mass of MgO: Mass of MgO = Moles of MgO × Formula weight of MgO = 0.02703 mol × 40.31 g/mol = 1.090 g

Step 5: (e) Determine moles and mass (g) of TiO₂ required

From the balanced chemical reaction, 1 mole of MgO reacts with 1 mole of TiO₂ to form 1 mole of MgTiO₃. So to form 0.02703 mol of MgTiO₃, 0.02703 mol of TiO₂ is required. To find the mass of TiO₂: Mass of TiO₂ = Moles of TiO₂ × Formula weight of TiO₂ = 0.02703 mol × 79.87 g/mol = 2.160 g To summarize, you will require 1.090 g of MgO and 2.160 g of TiO₂ to prepare 3.250 g of MgTiO₃.

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, the reactants, transform into different substances called products. In this exercise, the reaction between magnesium oxide (MgO) and titanium dioxide (TiO₂) produces magnesium titanate (MgTiO₃). Here, MgO and TiO₂ are the reactants, and MgTiO₃ is the product. The equation of this chemical reaction is:

• MgO + TiO₂ → MgTiO₃

This equation demonstrates a balanced chemical reaction, meaning that the number of atoms for each element is the same on both sides of the equation.
Balancing chemical equations is crucial because it complies with the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction.
Understanding how to read and balance chemical reactions is essential for predicting the outcome of a reaction and for calculating how much starting material is required or how much product will be formed.

Moles and Stoichiometry

Moles and stoichiometry are fundamental concepts in understanding chemical reactions on a quantitative level. A mole is a unit used for counting atoms, molecules, or particles. One mole corresponds to Avogadro's number, which is approximately 6.022 x 10²³ particles.
In this exercise, the stoichiometry of the balanced equation tells us the relationship between the reactants and the product in terms of moles. From the balanced equation, we see that 1 mole of MgO reacts with 1 mole of TiO₂ to produce 1 mole of MgTiO₃.
Using stoichiometry, we can calculate the moles needed for each reactant to make a certain amount of product. In this scenario, the goal was to produce 3.250 g of MgTiO₃. By knowing the formula weight of MgTiO₃ (120.18 g/mol), the moles of MgTiO₃ can be computed as follows:

• Moles of MgTiO₃ = \( \frac{3.250 \text{ g}}{120.18 \text{ g/mol}} \approx 0.02703 \text{ mol} \)

With this information, you can determine that 0.02703 moles of MgO and 0.02703 moles of TiO₂ are required to produce the MgTiO₃.

Molecular Weight Calculations

Molecular weight calculations involve determining the weight of a chemical compound based on its elemental composition, which is important for converting between grams and moles. The molecular weight is the sum of the atomic weights of all atoms in a compound.
To calculate the molecular weight of magnesium titanate (MgTiO₃), you add up the atomic weights of magnesium (Mg), titanium (Ti), and oxygen (O). Here are the atomic weights:

• Mg: 24.31 g/mol
• Ti: 47.87 g/mol
• O: 16.00 g/mol

Consequently, the formula for MgTiO₃ is calculated as:

• MgTiO₃ = 24.31 + 47.87 + 3(16.00) = 120.18 g/mol

This molecular weight allows us to convert the mass of a compound into moles, which is essential for using stoichiometry to find how much reactant is needed or how much product is formed in a chemical reaction.