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Chapter 3: Problem 48

What number of Fe atoms and what amount (moles) of Fe atoms are in 500.0 g of iron?

Short Answer

Expert verified
In a 500.0 g sample of iron, there are \(8.952\) moles of iron (Fe) atoms and approximately \(5.389 \times 10^{24}\) iron (Fe) atoms.

Step by step solution

01

Calculate the moles of iron (Fe)

To find the moles of iron, we will use the given mass (500.0 g) and divide it by the molar mass of iron (55.845 g/mol). This will give us the amount (moles) of Fe atoms in the sample. Moles of Fe = (mass of iron) / (molar mass of iron) Moles of Fe = (500.0 g) / (55.845 g/mol) Moles of Fe = 8.952 moles
02

Calculate the number of iron atoms (Fe)

To find the number of iron atoms in the sample, we will use the moles of iron (from Step 1) and multiply it by Avogadro's number (6.022 x 10^23 atoms/mol). Number of Fe atoms = (moles of Fe) x (Avogadro's number) Number of Fe atoms = (8.952 moles) x (6.022 x 10^23 atoms/mol) Number of Fe atoms = 5.389 x 10^24 atoms In conclusion, there are \(8.952\) moles of iron (Fe) atoms and approximately \(5.389 \times 10^{24}\) iron (Fe) atoms in 500.0 g of iron.

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Most popular questions from this chapter

Anabolic steroids are performance enhancement drugs whose use has been banned from most major sporting activities. One anabolic steroid is fluoxymesterone \(\left(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\right)\) . Calculate the percent composition by mass of fluoxymesterone.

In using a mass spectrometer, a chemist sees a peak at a mass of 30.0106 . Of the choices $^{12} \mathrm{C}_{2}^{1} \mathrm{H}_{6},^{12} \mathrm{C}^{1} \mathrm{H}_{2}^{16} \mathrm{O},\( and \)^{14} \mathrm{N}^{16} \mathrm{O}$ which is responsible for this peak? Pertinent masses are \(^{1} \mathrm{H}\) $1.007825 ; 16 \mathrm{O}, 15.994915 ;\( and \)^{14} \mathrm{N}, 14.003074$

Natural rubidium has the average mass of 85.4678 \(\mathrm{u}\) and is composed of isotopes \(^{85} \mathrm{Rb}(\mathrm{mass}=84.9117 \mathrm{u})\) and $^{87} \mathrm{Rb}\( . The ratio of atoms \)^{85} \mathrm{Rb} /^{87} \mathrm{Rb}$ in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\) .

Arrange the following substances in order of increasing mass percent of nitrogen. $\begin{array}{ll}{\text { a. } \mathrm{NO}} & {\text { c. } \mathrm{NH}_{3}} \\\ {\text { b. } \mathrm{N}_{2} \mathrm{O}} & {\text { d. SNH }}\end{array}$

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce $1.0 \times 10^{6} \mathrm{kg}\( \)\mathrm{HNO}_{3}\( by the Ostwald process? Assume 100\)\%$ yield in each reaction, and assume that the NO produced in the third step is not recycled.
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