Fluorine reacts with uranium hexafluoride, \(\mathrm{UF}_{6}\), as represented by this equation: \(\mathrm{U}(\mathrm{s})+3 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{UF}_{6}(\mathrm{~g})\) How many fluorine molecules are required to produce \(2.0 \mathrm{mg}\) of uranium hexafluoride, \(\mathrm{UF}_{6}\), from an excess of uranium? The molar mass of \(\mathrm{UF}_{6}\) is \(352.0 \mathrm{~g} \mathrm{~mol}^{-1} .\) (a) \(5.13 \times 10^{18}\) (b) \(1.026 \times 10^{19}\) (c) \(2.052 \times 10^{19}\) (d) \(1.026 \times 10^{20}\)

Short Answer

Expert verified
Approximately \(1.026 \times 10^{19}\) fluorine molecules are required.

Step by step solution

01

Convert mass of UF6 to moles

To determine the number of fluorine molecules required, first convert the mass of UF6 from milligrams to grams. Then use the molar mass of UF6 to find the number of moles of UF6.
02

Use the chemical equation

The balanced chemical equation indicates that 1 mole of UF6 is produced from 3 moles of F2. Use the stoichiometry of the reaction to find the moles of F2 needed.
03

Calculate the number of F2 molecules

To find the number of F2 molecules, multiply the moles of F2 by Avogadro's number (approximately 6.022 \(\times\) 10^23 molecules/mol).

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Key Concepts

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

Chemical Reaction Stoichiometry
Chemical reaction stoichiometry is a branch of chemistry that quantifies the relationships between reactants and products in chemical reactions. Using stoichiometry, we can predict how much of each substance is needed or produced in a reaction. The key tool for this prediction is the balanced chemical equation, which provides the mole-to-mole ratio between reactants and products.

Consider a simple reaction where substance A reacts with substance B to produce substance C. If the balanced equation states that 1 molecule of A reacts with 2 molecules of B to produce 3 molecules of C, we have a clear relationship: for every mole of A, we need two moles of B to make three moles of C. This example illustrates the fundamental principle of stoichiometry: the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. Therefore, the amount of reactants consumed will equal the amount of products formed, allowing us to perform exact calculations.
Mole-to-Mole Ratio Calculations
Mole-to-mole ratio calculations are a critical part of solving stoichiometry problems. These ratios come directly from the coefficients of a balanced chemical equation. The coefficients indicate the number of moles of each substance involved in the reaction.

In the case of the reaction between uranium (U) and fluorine (F2), the balanced equation tells us that 1 mole of U reacts with 3 moles of F2 to produce 1 mole of UF6. To calculate the amount of a reactant required or the amount of a product yielded, we need to compare the given information to these ratios. For example, if we are given the mass of the product UF6 and need to find the number of F2 molecules needed, we first convert the mass of UF6 to moles, and then use the 1:3 mole ratio to find the moles of F2 required.
Using Avogadro's Number
Avogadro's number, approximately 6.022 x 10^23 molecules/mol, is a fundamental constant in chemistry that represents the number of units (atoms, molecules, ions, etc.) in one mole of any substance. It bridges the gap between the atomic scale and the macroscopic scale, allowing chemists to work with quantities of substances in the lab.

To utilize Avogadro's number in stoichiometry problems, after calculating the moles of a reactant or product needed, we multiply that amount by Avogadro's number to convert it to the actual number of molecules or atoms. When asked how many fluorine molecules are required to produce a certain amount of UF6, after finding the moles of F2 needed using the balanced chemical equation and mole-to-mole ratios, we then multiply by Avogadro's number to find the precise number of F2 molecules.

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Most popular questions from this chapter

An ore contains \(2.296 \%\) of the mineral argentite, \(\mathrm{Ag}_{2} \mathrm{~S}\), by mass. How many grams of this ore would have to be processed in order to obtain \(1.00 \mathrm{~g}\) of pure solid silver? \((\mathrm{Ag}=108)\) (a) \(1.148 \mathrm{~g}\) (b) \(0.026 \mathrm{~g}\) (c) \(50 \mathrm{~g}\) (d) \(2.296 \mathrm{~g}\)

A sample of pure Cu \((4.00 \mathrm{~g})\) heated in a stream of oxygen for some time, gains in weight with the formation of black oxide of copper \((\mathrm{CuO})\). The final mass is 4.90 g. What per cent of copper remains unoxidized? \((\mathrm{Cu}=64)\) (a) \(90 \%\) (b) \(10 \%\) (c) \(20 \%\) (d) \(80 \%\)

Two elements \(\mathrm{A}\) and \(\mathrm{B}\) combine to form compound \(X\) and \(Y\). For the fix mass of A, masses of \(\mathrm{B}\) combined for the compounds \(\mathrm{A}\) and \(\mathrm{B}\) are in \(3: 7\) ratio. If in compound \(X, 4 g\) of \(A\) combines with \(12 \mathrm{~g} \mathrm{~B}\), then in compound \(\mathrm{Y}, 8 \mathrm{~g}\) of \(\mathrm{A}\) will combine with \(\ldots \ldots \mathrm{g}\) of \(\mathrm{B}\). (a) 24 (b) 56 (c) 28 (d) 8

A pre-weighed vessel was filled with oxygen at NTP and weighed. It was then evacuated, filled with \(\mathrm{SO}_{2}\) at the same temperature and pressure, and again weighed. The weight of oxygen is (a) the same as that of \(\mathrm{SO}_{2}\) (b) \(\frac{1}{2}\) that of \(\mathrm{SO}_{2}\) (c) twice that of \(\mathrm{SO}_{2}\) (d) \(\frac{1}{4}\) that of \(\mathrm{SO}_{2}\)

What is the total mass of the products formed, when \(51 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) is oxidized by oxygen to produce water and sulphur dioxide? (a) \(72 \mathrm{~g}\) (b) \(27 \mathrm{~g}\) (c) \(123 \mathrm{~g}\) (d) \(96 \mathrm{~g}\)

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