An iron bar weighed \(664 \mathrm{~g}\). After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right) .\) Calculate the final mass of the iron bar and rust.

Short Answer

Expert verified
The final mass of the iron bar and rust is approximately 699.8 g.

Step by step solution

01

Calculate the Mass of Iron that Turned into Rust

Since one-eighth of the iron turned into rust, you should divide the initial mass of the iron bar by 8. Hence, the mass of the iron that turned into rust is \( \frac{664g}{8} = 83g \).
02

Find the Mass of the Resulted Rust

The equation of rust is given by \(4Fe + 3O_2 = 2Fe_2O_3\). From this equation, we can calculate the amount of rust formed from a given mass of iron. According to molecular masses: \(4mol\ of\ Fe = 2mol\ of\ Fe_2O_3\). Therefore, \(1 mol\ of\ Fe = 0.5 mol\ of\ Fe_2O_3\). Now, calculate the number of molds of \(Fe\) which turned into rust: number of molds of \(Fe = \frac{mass}{molar\ mass} = \frac{83g}{55.845g/mol} = about\ 1.487mol\). Consequently, the number of moulds of \(Fe_2O_3\) formed is \(0.5*1.487 = 0.7435mol\). Finally, the mass of the rust formed is \(molar\ mass * number\ of\ moles = 159.687g/mol * 0.7435 mol = about 118.8g\).
03

Calculate Final Mass of Iron Bar and Rust

To calculate the final mass of the iron bar and rust, start with the initial mass of the iron bar (664g), then subtract the mass of the iron that turned to rust (83g), and add the mass of the rust that formed (approximately 118.8g). Hence, the final mass is composed of the mass of the iron and the rust: \(664g - 83g + 118.8g = approximately 699.8g\).

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes in which substances, known as reactants, transform into new substances called products. These transformations occur according to certain rules of chemical combination, dictated by factors such as the conservation of mass and stoichiometric ratio. The reaction we're investigating involves iron and oxygen, which react together to form rust.

Diving into the details, the balanced chemical equation \(4Fe + 3O_2 = 2Fe_2O_3\) provides essential information for stoichiometry calculations. Each molecule participates in the reaction with specific proportions, with 4 moles of iron reacting with 3 moles of oxygen to yield 2 moles of iron(III) oxide (rust). This stoichiometric relationship helps us understand how much of each reactant is needed and the amount of product that will be formed.
Rust Formation
Rust formation is a common chemical process, also known as oxidation, where iron reacts with water and oxygen in the environment to form iron oxide. The specifics of the reaction can vary depending on conditions like humidity and temperature, but the principle remains constant: iron atoms lose electrons to oxygen atoms, resulting in a compound commonly known as rust or \(Fe_2O_3\).

From a practical standpoint, rust is undesirable as it weakens metallic structures. In our exercise, one-eighth of the iron bar has rusted over time in moist air. This detail is critical because it gives us a quantifiable value to plug into our stoichiometry calculations, allowing us to estimate the weight of the rust and the remaining iron bar accurately.
Mole Concept
The mole concept is a fundamental principle in chemistry that relates the mass of a substance to the number of its particles, such as atoms, molecules, or ions. One mole of any substance contains Avogadro's number (approx. \(6.022 \times 10^{23}\)) of particles, and its mass is equivalent to the substance's molar mass expressed in grams.

Applying the mole concept in our exercise, we convert the mass of iron that has turned to rust into moles to find out how many moles of \(Fe_2O_3\) are formed. This step is pivotal because it bridges the gap between the mass of iron and the final mass of rust, which are not direct 1:1 conversions due to the involvement of oxygen atoms from the air in the rust compound.

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Most popular questions from this chapter

Define limiting reactant and excess reactant. What is the significance of the limiting reactant in predicting the amount of the product obtained in a reaction? Can there be a limiting reactant if only one reactant is present?

Industrially, nitric acid is produced by the Ostwald process represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\\2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{HNO}_{3}(a q)+\mathrm{HNO}_{2}(a q) \end{aligned}$$What mass of \(\mathrm{NH}_{3}\) (in grams) must be used to produce 1.00 ton of \(\mathrm{HNO}_{3}\) by the above procedure, assuming an 80 percent yield in each step? ( 1 ton = \(2000 \mathrm{lb} ; 1 \mathrm{lb}=453.6 \mathrm{~g} .)\)

Which of the following has the greater mass: \(0.72 \mathrm{~g}\) of \(\mathrm{O}_{2}\) or 0.0011 mole of chlorophyll \(\left(\mathrm{C}_{55} \mathrm{H}_{72} \mathrm{MgN}_{4} \mathrm{O}_{5}\right) ?\)

An impure sample of zinc (Zn) is treated with an excess of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) to form zinc sulfate \(\left(\mathrm{ZnSO}_{4}\right)\) and molecular hydrogen \(\left(\mathrm{H}_{2}\right) .\) (a) Write a balanced equation for the reaction. (b) If \(0.0764 \mathrm{~g}\) of \(\mathrm{H}_{2}\) is obtained from \(3.86 \mathrm{~g}\) of the sample, calculate the percent purity of the sample. (c) What assumptions must you make in (b)?

Balance the following equations using the method outlined in Section 3.7 . (a) \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}\) (b) \(\mathrm{KNO}_{3} \longrightarrow \mathrm{KNO}_{2}+\mathrm{O}_{2}\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\) (e) \(\mathrm{NaHCO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) (f) \(\mathrm{P}_{4} \mathrm{O}_{10}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}\) (g) \(\mathrm{HCl}+\mathrm{CaCO}_{3} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) (h) \(\mathrm{Al}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2}\) (i) \(\mathrm{CO}_{2}+\mathrm{KOH} \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}\) (j) \(\mathrm{CH}_{4}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (k) \(\mathrm{Be}_{2} \mathrm{C}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Be}(\mathrm{OH})_{2}+\mathrm{CH}_{4}\) (l) \(\mathrm{Cu}+\mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\) \((\mathrm{m}) \mathrm{S}+\mathrm{HNO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (n) \(\mathrm{NH}_{3}+\mathrm{CuO} \longrightarrow \mathrm{Cu}+\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\)

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