/*! 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 49 Aluminum metal is produced by pa... [FREE SOLUTION] | 91Ó°ÊÓ

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Aluminum metal is produced by passing an electric current through a solution of aluminum oxide \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) dissolved in molten cryolite \(\left(\mathrm{Na}_{3} \mathrm{AlF}_{6}\right) .\) Calculate the molar masses of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) and \(\mathrm{Na}_{3} \mathrm{AlF}_{6}.\)

Short Answer

Expert verified
The molar masses of Aluminum Oxide (Al₂O₃) and Cryolite (Na₃AlF₆) are 102 g/mol and 210 g/mol, respectively.

Step by step solution

01

Atomic masses of elements

Look up the atomic masses of each element in the periodic table, rounded to the nearest whole number: - Aluminum (Al): 27 g/mol - Oxygen (O): 16 g/mol - Sodium (Na): 23 g/mol - Fluorine (F): 19 g/mol
02

Calculate the molar mass of Al₂O₃

First, we need to calculate the molar mass for Aluminium Oxide: - 2 moles of Al = 2 × 27 g/mol - 3 moles of O = 3 × 16 g/mol Now, add these values together (2 × 27) + (3 × 16) = 54 + 48 = 102 g/mol So, the molar mass of Al₂O₃ is 102 g/mol.
03

Calculate the molar mass of Na₃AlF₆

Next, we'll calculate the molar mass for Cryolite: - 3 moles of Na = 3 × 23 g/mol - 1 mole of Al = 1 × 27 g/mol - 6 moles of F = 6 × 19 g/mol Now, add these values together (3 × 23) + (1 × 27) + (6 × 19) = 69 + 27 + 114 = 210 g/mol So, the molar mass of Na₃AlF₆ is 210 g/mol. As a result, the molar masses are as follows: - Aluminium Oxide (Al₂O₃): 102 g/mol - Cryolite (Na₃AlF₆): 210 g/mol

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Key Concepts

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

Aluminium Oxide
Aluminium oxide, represented as \( \mathrm{Al}_{2}\mathrm{O}_{3} \) in chemical notation, is a common compound formed from aluminium and oxygen. It occurs naturally as corundum or bauxite and has various industrial applications, including being used as an abrasive and as a component in the production of aluminum metal.

The molar mass of aluminium oxide is a concept that combines the individual atomic masses of aluminum and oxygen. To calculate this, we simply multiply the atomic mass of each element by its stoichiometric coefficient in the formula, which represents the number of atoms of the element in one molecule of the substance. For \( \mathrm{Al}_{2}\mathrm{O}_{3} \) the calculation is straightforward: With 2 atoms of aluminum and 3 atoms of oxygen, we determine the molar mass by adding twice the atomic mass of aluminum to thrice the atomic mass of oxygen, resulting in a molar mass of 102 g/mol. Understanding the molar mass is crucial when dealing with reactions involving aluminium oxide to measure or predict the amounts of reactants and products efficiently.
Cryolite
Cryolite, or \( \mathrm{Na}_{3}\mathrm{AlF}_{6} \) is a less common mineral used in the metallurgy of aluminium. It serves as a solvent for aluminium oxide in the Hall-Héroult process, which is essential for extracting aluminium from its oxide ore. The role of cryolite is to lower the melting point of aluminium oxide, making the extraction process more energy-efficient.

When calculating the molar mass of cryolite, it is important to note that the substance is made up of sodium, aluminium, and fluorine. Just as with aluminium oxide, we calculate its molar mass by multiplying the atomic mass of each constituent element by the number of atoms present in one mole of cryolite, and then summing these together. This results in a calculated molar mass of 210 g/mol for cryolite. Knowledge of its molar mass is valuable in the field of materials science and when performing quantitative analyses of materials containing cryolite.
Atomic Masses
Atomic masses, a fundamental value found on the periodic table, reflect the mass of a singular atom of a chemical element measured in atomic mass units (amu), which is nearly equivalent to the mass in grams per mole of the element. To craft molar mass calculations, one needs accurate atomic masses.

By peeking at the periodic table, we can glean the atomic masses for aluminium, oxygen, sodium, and fluorine—essential components in our earlier calculations. Through multiplying these atomic masses by the corresponding number of atoms in a molecule and summing up the products, we arrive at the molar mass. For students and chemists alike, understanding and utilizing atomic masses is indispensable for converting between moles and grams, a common task in chemistry.
Stoichiometry
Stoichiometry is the section of chemistry focusing on the quantitative relationships between the amounts of reactants and products in a chemical reaction. It's grounded in the conservation of mass and the law of definite proportions.

When we carry out molar mass calculations involving complex compounds such as aluminium oxide and cryolite, stoichiometry helps us determine the exact proportions of each element necessary to form these compounds. It explains why we multiply the atomic masses of aluminium, oxygen, sodium, and fluorine by 2, 3, 3, and 6, respectively, in the formulas \( \mathrm{Al}_{2}\mathrm{O}_{3} \) and \( \mathrm{Na}_{3}\mathrm{AlF}_{6} \). These coefficients are stoichiometric coefficients indicating the number of moles of each element in one mole of the compound. Stoichiometry is not just theoretical; it’s the tool chemists use to make predictions about reaction outcomes and to calculate necessary reagent quantities accurately.
Periodic Table
The periodic table is not merely a chart but the gateway to understanding elemental characteristics and their relationships. It's organized in such a manner that elements with similar properties fall into the same columns, known as groups, and elements with increasing atomic number line up in rows, referred to as periods.

The periodic table also provides essential data like atomic masses, which we refer to when calculating molar masses. It's the go-to resource for any chemist when they need to ascertain the atomic mass of an element, as seen in the computations for \( \mathrm{Al}_{2}\mathrm{O}_{3} \) and \( \mathrm{Na}_{3}\mathrm{AlF}_{6}\). For students, the periodic table is crucial as it assists in visualizing the framework on which the atomic world operates, equipping them to grasp more advanced concepts in chemistry with confidence.

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Most popular questions from this chapter

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound. If \(25.0 \mathrm{g} \mathrm{Ag}_{2} \mathrm{O}\) is reacted with \(50.0 \mathrm{g}\) \(\mathrm{C}_{10} \mathrm{H}_{10} \mathrm{N}_{4} \mathrm{SO}_{2},\) what mass of silver sulfadiazine, \(\mathrm{AgC}_{10} \mathrm{H}_{9} \mathrm{N}_{4} \mathrm{SO}_{2},\) can be produced, assuming \(100 \%\) yield? $$\mathrm{Ag}_{2} \mathrm{O}(s)+2 \mathrm{C}_{10} \mathrm{H}_{10} \mathrm{N}_{4} \mathrm{SO}_{2}(s) \longrightarrow 2 \mathrm{AgC}_{10} \mathrm{H}_{9} \mathrm{N}_{4} \mathrm{SO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)$$

Adipic acid is an organic compound composed of \(49.31 \%\) C, \(43.79 \% \mathrm{O},\) and the rest hydrogen. If the molar mass of adipic acid is \(146.1 \mathrm{g} / \mathrm{mol},\) what are the empirical and molecular formulas for adipic acid?

A common demonstration in chemistry courses involves adding a tiny speck of manganese(IV) oxide to a concentrated hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) solution. Hydrogen peroxide decomposes quite spectacularly under these conditions to produce oxygen gas and steam (water vapor). Manganese(IV) oxide is a catalyst for the decomposition of hydrogen peroxide and is not consumed in the reaction. Write the balanced equation for the decomposition reaction of hydrogen peroxide.

The reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas produces \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) as its main product (along with HCl). In addition, the reaction invariably produces a variety of other minor products, including \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3},\) and others. Naturally, the production of these minor products reduces the yield of the main product. Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction of \(300 .\) g of ethane with \(650 .\) g of chlorine produced \(490 .\) g of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}.\)

With the advent of techniques such as scanning tunneling microscopy, it is now possible to "write" with individual atoms by manipulating and arranging atoms on an atomic surface. a. If an image is prepared by manipulating iron atoms and their total mass is \(1.05 \times 10^{-20} \mathrm{g},\) what number of iron atoms were used? b. If the image is prepared on a platinum surface that is exactly 20 platinum atoms high and 14 platinum atoms wide, what is the mass (grams) of the atomic surface? c. If the atomic surface were changed to ruthenium atoms and the same surface mass as determined in part b is used, what number of ruthenium atoms is needed to construct the surface?

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