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Ch.6 Chemical Reactions: Mole and Mass Relationships
McMurry - Fundamentals of GOB 8th Edition
McMurry8th EditionFundamentals of GOBISBN: 9780134015187Not the one you use?Change textbook
Chapter 6, Problem 43c

Titanium metal is obtained from the mineral rutile, TiO2. The process requires multiple steps, as shown in the following reactions:
TiO2(s) + 2 Cl2(g) + 2 C(s) → TiCl4(s) + 2 CO(g)
TiCl4(s) + 2 Mg(s) → Ti(s) + 2 MgCl2(s)
c. How many kilograms of rutile are needed to produce 95 kg of Ti?

Verified step by step guidance
1
Step 1: Write down the balanced chemical equations provided in the problem. The first reaction is: TiO₂(s) + 2 Cl₂(g) + 2 C(s) → TiCl₄(s) + 2 CO(g). The second reaction is: TiCl₄(s) + 2 Mg(s) → Ti(s) + 2 MgCl₂(s). These equations show the conversion of TiO₂ to TiCl₄ and then to Ti.
Step 2: Determine the molar masses of the relevant compounds. Use the periodic table to calculate the molar masses: TiO₂ (47.87 g/mol for Ti + 2 × 16.00 g/mol for O), Ti (47.87 g/mol), and any other compounds as needed.
Step 3: Convert the given mass of titanium (95 kg) to moles. Use the molar mass of titanium (Ti) to perform the conversion: moles of Ti = mass of Ti (in grams) ÷ molar mass of Ti.
Step 4: Use stoichiometry to determine the moles of TiO₂ required. From the first reaction, 1 mole of TiO₂ produces 1 mole of TiCl₄, and from the second reaction, 1 mole of TiCl₄ produces 1 mole of Ti. Therefore, the moles of TiO₂ required are equal to the moles of Ti calculated in Step 3.
Step 5: Convert the moles of TiO₂ to mass in kilograms. Use the molar mass of TiO₂ to calculate the mass: mass of TiO₂ = moles of TiO₂ × molar mass of TiO₂. Finally, convert the result to kilograms by dividing by 1000.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to calculate the amounts of substances consumed and produced in a reaction based on balanced chemical equations. Understanding stoichiometry is essential for determining how much rutile is needed to produce a specific mass of titanium.
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Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is crucial for converting between the mass of a substance and the number of moles, which is necessary for stoichiometric calculations. For this problem, knowing the molar masses of TiO2 and Ti will help in determining how much rutile is required to yield the desired amount of titanium.
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Chemical Reactions

Chemical reactions involve the transformation of reactants into products through breaking and forming chemical bonds. In this case, the reactions show how titanium is extracted from rutile via chlorination and reduction. Understanding the steps and the stoichiometric coefficients in these reactions is vital for calculating the amount of rutile needed to produce a specific quantity of titanium.
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Related Practice
Textbook Question

The principal component of many kidney stones is calcium oxalate, CaC2O4. A kidney stone recovered from a typical patient contains 8.5 × 1020 formula units of calcium oxalate. How many moles of CaC2O4 are present in this kidney stone? What is the mass of the kidney stone in grams?

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Textbook Question

Titanium metal is obtained from the mineral rutile, TiO2. The process requires multiple steps, as shown in the following reactions:

TiO2(s) + 2 Cl2(g) + 2 C(s) → TiCl4(s) + 2 CO(g)

TiCl4(s) + 2 Mg(s) → Ti(s) + 2 MgCl2(s)

a. Write mole ratios to show the relationship between the reactants and products for each reaction.

1808
views
Textbook Question

Titanium metal is obtained from the mineral rutile, TiO2. The process requires multiple steps, as shown in the following reactions:

TiO2(s) + 2 Cl2(g) + 2 C(s) → TiCl4(s) + 2 CO(g)

TiCl4(s) + 2 Mg(s) → Ti(s) + 2 MgCl2(s)

b. How many moles of TiO2 are needed to form one mole of titanium?

1870
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Textbook Question

In Problem 6.40, hydrazine reacted with oxygen according to the following (unbalanced) equation: N2H4(l) + O2(g) → NO2(g) + H2O(g)

a. If 75.0 kg of hydrazine are reacted with 75.0 kg of oxygen, which is the limiting reagent?

1650
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Textbook Question

Nitrobenzene (C6H5NO2) is used in small quantities as a flavoring agent or in perfumes but can be toxic in large amounts. It is produced by reaction of benzene (C6H6) with nitric acid:

C6H6(l) + HNO3(aq) → C6H5NO2(l) + H2O(l)

a. Identify the limiting reagent in the reaction of 27.5 g of nitric acid with 75 g of benzene.

2148
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Textbook Question

Nitrobenzene (C6H5NO2) is used in small quantities as a flavoring agent or in perfumes but can be toxic in large amounts. It is produced by reaction of benzene (C6H6) with nitric acid:

C6H6(l) + HNO3(aq) → C6H5NO2(l) + H2O(l)

b. Calculate the theoretical yield for this reaction.

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