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Chapter 12: Problem 131

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.

Short Answer

Expert verified
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 by step solution

01

(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.
02

(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
03

(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
04

(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
05

(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.

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Indicate whether each statement is true or false: (a) The band gap of a semiconductor decreases as the particle size decreases in the \(1-10-\mathrm{nm}\) range. (b) The light that is emitted from a semiconductor, upon external stimulation, becomes longer in wavelength as the particle size of the semiconductor decreases.

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