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Chapter 3: Problem 96

Titanium(IV) oxide \(\left(\mathrm{TiO}_{2}\right)\) is a white substance produced by the action of sulfuric acid on the mineral ilmenite \(\left(\mathrm{FeTiO}_{3}\right):\) $$ \mathrm{FeTiO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{TiO}_{2}+\mathrm{FeSO}_{4}+\mathrm{H}_{2} \mathrm{O} $$ Its opaque and nontoxic properties make it suitable as a pigment in plastics and paints. In one process, \(8.00 \times\) \(10^{3} \mathrm{~kg}\) of \(\mathrm{FeTiO}_{3}\) yielded \(3.67 \times 10^{3} \mathrm{~kg}\) of \(\mathrm{TiO}_{2}\). What is the percent yield of the reaction?

Short Answer

Expert verified
The percent yield of the reaction is 86.98%.

Step by step solution

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01

Understanding the Chemical Equation

The given chemical reaction is \(\text{FeTiO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{TiO}_2 + \text{FeSO}_4 + \text{H}_2\text{O}\). From this, we can determine that 1 mole of \(\text{FeTiO}_3\) reacts to produce 1 mole of \(\text{TiO}_2\).
02

Calculate Moles of FeTiO3

To find the moles of \(\text{FeTiO}_3\), we first need its molar mass: \(\text{Fe} = 55.85\), \(\text{Ti} = 47.87\), and \(\text{O} = 16.00\) u. Thus, \(\text{FeTiO}_3\) has a molar mass of \(55.85 + 47.87 + 3 \times 16 = 151.72\) g/mol. We have \(8.00 \times 10^3\) kg of \(\text{FeTiO}_3\), which is \(8.00 \times 10^6\) g. Calculate the moles: \(\frac{8.00 \times 10^6 \text{ g}}{151.72 \text{ g/mol}} = 52758.44\text{ mol}\).
03

Calculate Theoretical Yield of TiO2

According to the equation, 1 mole of \(\text{FeTiO}_3\) produces 1 mole of \(\text{TiO}_2\). Therefore, theoretically, \(52758.44\) moles of \(\text{FeTiO}_3\) will produce \(52758.44\) moles of \(\text{TiO}_2\). With \(\text{TiO}_2\)'s molar mass being \(47.87 + 2 \times 16 = 79.87\) g/mol, the mass is calculated as: \(52758.44\text{ mol}\times 79.87 \text{ g/mol} = 4217006.51\text{ g}\), or \(4217.01 \text{ kg}\).
04

Calculate Percent Yield

The actual yield of \(\text{TiO}_2\) is \(3.67 \times 10^3\) kg. The percent yield is given by: \(\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\). Substitute the values: \(\frac{3.67 \times 10^3 \text{ kg}}{4217.01 \text{ kg}} \times 100 = 86.98\%\).

Key Concepts

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

Chemical Reaction
A chemical reaction describes the process in which atoms are rearranged to form new substances. In this case, ilmenite (\(\text{FeTiO}_3\)) reacts with sulfuric acid (\(\text{H}_2\text{SO}_4\)) to produce titanium(IV) oxide (\(\text{TiO}_2\)), iron(II) sulfate (\(\text{FeSO}_4\)), and water (\(\text{H}_2\text{O}\)). It is important to recognize that chemical reactions adhere to the law of conservation of mass, meaning that the total mass of reactants equals the total mass of products. This is evident in balanced chemical equations, which show the same number of each type of atom on both sides of the reaction.
In the equation provided, the states of matter and coefficients indicate that for every mole of \(\text{FeTiO}_3\) consumed, one mole of \(\text{TiO}_2\) is produced. Understanding this stoichiometry is crucial for calculating yields from a reaction.
Theoretical Yield
Theoretical yield refers to the maximum amount of product that could be formed from a given amount of reactants, based on the stoichiometry of the chemical reaction. This assumes that the reaction goes to completion with no side reactions or losses.
From the equation provided: \(\text{FeTiO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{TiO}_2 + \text{FeSO}_4 + \text{H}_2\text{O}\), one mole of \(\text{FeTiO}_3\) theoretically produces one mole of \(\text{TiO}_2\).
  • Using the molar mass of \(\text{FeTiO}_3\) (151.72 g/mol) and \(\text{TiO}_2\) (79.87 g/mol), we convert the mass of reactants to moles.
  • This allows us to calculate the moles of \(\text{TiO}_2\) theoretically produced.
  • The theoretical yield is then the mass of these moles.
In this scenario, the theoretical yield of \(\text{TiO}_2\) was found to be 4217.01 kg.
Actual Yield
Actual yield is the amount of product actually obtained from a chemical reaction, which may be lower than the theoretical yield due to factors such as incomplete reactions, side reactions, or loss of product during recovery.
It is essential to measure actual yield in laboratory or industrial processes to compare with theoretical yield.
In the problem given, the actual yield of \(\text{TiO}_2\) is 3.67 \(\times\) 10³ kg.
This yield can then be used to calculate the percent yield, which indicates the efficiency of the reaction or process.
Efforts in improving the actual yield often involve optimizing reaction conditions, purifying reactants, and efficient collection of the product, allowing for comparison between the potential maximum and the real-world results.
Molar Mass Calculation
Molar mass is the mass of one mole of a given substance, expressed in g/mol. To accurately calculate molar masses, you add together the atomic masses of all atoms in the molecular formula.
In the provided step-by-step solution:- The molar mass of \(\text{FeTiO}_3\) was calculated as:
Fe (55.85 g/mol) + Ti (47.87 g/mol) + 3O (3\(\times\)16.00 g/mol) = 151.72 g/mol.- The molar mass of \(\text{TiO}_2\) was determined using:
Ti (47.87 g/mol) + 2O (2\(\times\)16.00 g/mol) = 79.87 g/mol.Knowing the molar mass is crucial for converting between mass and moles during calculations.
  • It allows chemists to use the stoichiometric coefficients from the balanced equation to determine theoretical yield.
  • The calculation is foundational for further conversion and yield calculations in chemistry.
The conversion from grams to moles is essential for directly utilizing balanced equations in practical applications.

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Most popular questions from this chapter

The depletion of ozone \(\left(\mathrm{O}_{3}\right)\) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from high-altitude jet planes. The reaction is $$ \mathrm{O}_{3}+\mathrm{NO} \longrightarrow \mathrm{O}_{2}+\mathrm{NO}_{2} $$ If \(0.740 \mathrm{~g}\) of \(\mathrm{O}_{3}\) reacts with \(0.670 \mathrm{~g}\) of NO, how many grams of \(\mathrm{NO}_{2}\) will be produced? Which compound is the limiting reactant? Calculate the number of moles of the excess reactant remaining at the end of the reaction.

The following is a crude but effective method for estimating the order of magnitude of Avogadro's number using stearic acid \(\left(\mathrm{C}_{18} \mathrm{H}_{36} \mathrm{O}_{2}\right)\). When stearic acid is added to water, its molecules collect at the surface and form a monolayer; that is, the layer is only one molecule thick. The cross-sectional area of each stearic acid molecule has been measured to be \(0.21 \mathrm{nm}^{2}\). In one experiment, it is found that \(1.4 \times 10^{-4} \mathrm{~g}\) of stearic acid is needed to form a monolayer over water in a dish of diameter \(20 \mathrm{~cm}\). Based on these measurements, what is Avogadro's number? (The area of a circle of radius \(r\) is \(\pi r^{2}\).)

Ascorbic acid (vitamin C) contains \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O} . \mathrm{In}\) one combustion analysis, \(5.24 \mathrm{~g}\) of ascorbic acid yields \(7.86 \mathrm{~g} \mathrm{CO}_{2}\) and \(2.14 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) Calculate the empirical formula and molecular formula of ascorbic acid given that its molar mass is about \(176 \mathrm{~g}\).

A certain sample of coal contains 1.6 percent sulfur by mass. When the coal is burned, the sulfur is converted to sulfur dioxide. To prevent air pollution, this sulfur dioxide is treated with calcium oxide \((\mathrm{CaO})\) to form calcium sulfite \(\left(\mathrm{CaSO}_{3}\right) .\) Calculate the daily mass (in kilograms) of \(\mathrm{CaO}\) needed by a power plant that uses \(6.60 \times 10^{6} \mathrm{~kg}\) of coal per day.

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating Si in chlorine gas: $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) $$ In one reaction, \(0.507 \mathrm{~mol}\) of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?
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