/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Titanium dioxide \(\left(\mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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Titanium dioxide \(\left(\mathrm{TiO}_{2}\right)\), the substance used as the pigment in white paint, is prepared industrially by reaction of \(\mathrm{TiCl}_{4}\) with \(\mathrm{O}_{2}\) at high temperature: $$ \mathrm{TiCl}_{4}+\mathrm{O}_{2} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{TiO}_{2}+2 \mathrm{Cl}_{2} $$ How many kilograms of \(\mathrm{TiO}_{2}\) can be prepared from \(5.60 \mathrm{~kg}\) of \(\mathrm{TiCl}_{4} ?\)

Short Answer

Expert verified
The maximum amount of \( \mathrm{TiO}_{2} \) produced is 2.358 kg.

Step by step solution

01

Write a Balanced Equation

The balanced chemical equation for the reaction between \( \mathrm{TiCl}_{4} \) and \( \mathrm{O}_{2} \) is already provided as: \\[ \mathrm{TiCl}_{4} + \mathrm{O}_{2} \rightarrow \mathrm{TiO}_{2} + 2 \mathrm{Cl}_{2} \] \This equation indicates that one mole of \( \mathrm{TiCl}_{4} \) reacts with one mole of \( \mathrm{O}_{2} \) to form one mole of \( \mathrm{TiO}_{2} \).
02

Calculate Molar Masses

Calculate the molar masses of the compounds involved: \- \( \mathrm{TiCl}_{4} \): \( \text{Ti} = 47.87 \), \( \text{Cl} = 35.45 \times 4 = 141.8 \); Total = \( 47.87 + 141.8 = 189.67 \; \text{g/mol} \)\- \( \mathrm{TiO}_{2} \): \( \text{Ti} = 47.87 \), \( \text{O} = 16.00 \times 2 = 32.00 \); Total = \( 47.87 + 32.00 = 79.87 \; \text{g/mol} \)
03

Convert Mass to Moles

Convert the mass of \( 5.60 \; \text{kg} = 5600 \; \text{g} \) of \( \mathrm{TiCl}_{4} \) to moles using its molar mass: \\[ \text{Moles of } \mathrm{TiCl}_{4} = \frac{5600 \; \text{g}}{189.67 \; \text{g/mol}} \approx 29.53 \; \text{mol} \]
04

Use Stoichiometry to Find Moles of \( \mathrm{TiO}_{2} \)

From the balanced equation, the mole ratio of \( \mathrm{TiCl}_{4} \) to \( \mathrm{TiO}_{2} \) is 1:1. Therefore, the moles of \( \mathrm{TiO}_{2} \) formed are the same as the moles of \( \mathrm{TiCl}_{4} \), which is approximately 29.53 mol.
05

Convert Moles of \( \mathrm{TiO}_{2} \) to Mass in Kilograms

Calculate the mass of \( \mathrm{TiO}_{2} \) using its molar mass: \\[ \text{Mass of } \mathrm{TiO}_{2} = 29.53 \; \text{mol} \times 79.87 \; \text{g/mol} \approx 2358 \; \text{g} \]\Convert this to kilograms: \[ 2358 \; \text{g} = 2.358 \; \text{kg} \]
06

Conclusion: Result Interpretation

The maximum amount of \( \mathrm{TiO}_{2} \) that can be produced from 5.60 kg of \( \mathrm{TiCl}_{4} \) with excess \( \mathrm{O}_{2} \) is approximately 2.358 kg.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
Chemical reactions occur when substances, known as reactants, undergo a transformation to form new substances, called products. These reactions involve the breaking of bonds in the reactants and the formation of new bonds resulting in products. In the case of preparing titanium dioxide (TiOâ‚‚) from titanium tetrachloride (TiClâ‚„) and oxygen (Oâ‚‚), a chemical reaction takes place at high temperatures. The essence of chemical reactions lies in the rearrangement of atoms, leading to a change in the chemical composition of the substances involved. Every reaction is specific and depends on certain conditions, such as temperature, pressure, and the presence of catalysts, which can affect the rate and efficiency of the reaction.
Molar Mass
Molar mass is a critical concept in stoichiometry, which involves calculating the masses, moles, and numbers of atoms in a chemical reaction. It refers to the mass of one mole of a substance, measured in grams per mole (g/mol). Knowing the molar mass is necessary for converting between the mass of a substance and the amount in moles. In the reaction of TiClâ‚„ with Oâ‚‚ to produce TiOâ‚‚, calculating the molar mass helps in determining how much of each substance is needed or produced. For example:
  • The molar mass of TiClâ‚„ is calculated by summing the atomic masses of its constituent elements: Titanium (Ti) and Chlorine (Cl).
  • Similarly, the molar mass of TiOâ‚‚ is the sum of the atomic mass of Ti and twice that of Oxygen (O).
These calculations ensure accuracy when converting between grams and moles during the chemical process.
Balanced Chemical Equations
A balanced chemical equation accurately represents the number of atoms for each element in the reactants and products. Balancing chemical equations is essential because it obeys the law of conservation of mass. This means that the same number of atoms of each element must exist on both sides of the equation.In the synthesis of TiOâ‚‚ from TiClâ‚„ and Oâ‚‚, the balanced equation is given by:\[ \mathrm{TiCl}_{4} + \mathrm{O}_{2} \rightarrow \mathrm{TiO}_{2} + 2 \mathrm{Cl}_{2} \]This equation illustrates that one mole of TiClâ‚„ reacts with one mole of Oâ‚‚ to yield one mole of TiOâ‚‚ and two moles of Clâ‚‚.
  • Each component's atoms are tallied, confirming that the same number appear both in the reactants and products.
Balancing ensures that stoichiometric calculations can proceed correctly, helping determine the right amounts of substances involved.
Chemical Conversions
Chemical conversions involve changing one unit of measurement to another, such as converting mass to moles or moles back to mass. This is a vital part of stoichiometry, allowing chemists to predict how much of a substance is needed or produced in a chemical reaction.The process of converting between mass and moles is straightforward once the molar mass is known. For example, suppose you have 5.60 kg of TiClâ‚„. First, you convert kilograms to grams, turning 5.60 kg into 5600 g. Using the molar mass of TiClâ‚„ (189.67 g/mol), you calculate the number of moles by dividing the mass by the molar mass:\[ \text{Moles of } \mathrm{TiCl}_{4} = \frac{5600 \; \text{g}}{189.67 \; \text{g/mol}} \approx 29.53 \; \text{mol} \]Similarly, once you determine the moles of TiOâ‚‚ formed, you can find its mass using its molar mass. These conversions illustrate the flexibility and utility of stoichiometry in predicting and analyzing chemical reactions.

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Most popular questions from this chapter

Nickel(II) sulfate, used for nickel plating, is prepared by treatment of nickel(II) carbonate with sulfuric acid: $$ \mathrm{NiCO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{NiSO}_{4}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} $$ (a) How many grams of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to react with \(14.5 \mathrm{~g}\) of \(\mathrm{NiCO}_{3} ?\) (b) How many grams of \(\mathrm{NiSO}_{4}\) are obtained if the yield is \(78.9 \% ?\)

An unidentified metal M reacts with an unidentified halogen \(\mathrm{X}\) to form a compound \(\mathrm{MX}_{2}\). When heated, the compound decomposes by the reaction: $$ 2 \mathrm{MX}_{2}(s) \longrightarrow 2 \mathrm{MX}(s)+\mathrm{X}_{2}(g) $$ When \(1.12 \mathrm{~g}\) of \(\mathrm{MX}_{2}\) is heated, \(0.720 \mathrm{~g}\) of \(\mathrm{MX}\) is obtained, along with \(56.0 \mathrm{~mL}\) of \(\mathrm{X}_{2}\) gas. Under the conditions used, \(1.00 \mathrm{~mol}\) of the gas has a volume of \(22.41 \mathrm{~L}\). (a) What is the atomic weight and identity of the halogen \(\mathrm{X} ?\) (b) What is the atomic weight and identity of the metal M?

Calculate the formula weight or molecular weight of the following substances: (a) \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (rust) (c) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) (citric acid) (b) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (sulfuric acid) (d) \(\mathrm{C}_{16} \mathrm{H}_{18} \mathrm{~N}_{2} \mathrm{O}_{4} \mathrm{~S}\) (penicillin G)

Balance the following equations: (a) \(\mathrm{Mg}+\mathrm{HNO}_{3} \longrightarrow \mathrm{H}_{2}+\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{CaC}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{C}_{2} \mathrm{H}_{2}\) (c) \(\mathrm{S}+\mathrm{O}_{2} \longrightarrow \mathrm{SO}_{3}\) (d) \(\mathrm{UO}_{2}+\mathrm{HF} \longrightarrow \mathrm{UF}_{4}+\mathrm{H}_{2} \mathrm{O}\)

Pure oxygen was first made by heating mercury(II) oxide: \(\mathrm{Hg} \mathrm{O} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{Hg}+\mathrm{O}_{2} \quad\) Unbalanced (a) Balance the equation. (b) How many grams of mercury and how many grams of oxygen are formed from \(45.5 \mathrm{~g}\) of \(\mathrm{HgO}\) ? (c) How many grams of \(\mathrm{HgO}\) would you need to obtain \(33.3 \mathrm{~g}\) of \(\mathrm{O}_{2} ?\)

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