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Chapter 3: Problem 99

An iron bar weighed 664 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 the rust is 818.96 g.

Step by step solution

01

Find the weight of the iron that turned into rust

The initial weight of the iron bar is 664 g. As one-eighth of the iron turned into rust, the weight of the iron that rusted is: \( \frac{1}{8} \times 664 \, g = 83 \, g \). This is the weight of the iron (Fe) that turned into rust (Fe2O3).
02

Calculate the weight of rust formed

To convert the weight of iron (Fe) that turned into rust (Fe2O3) to the weight of the rust formed, we need to multiply by the ratio of the molecular weight of rust to the molecular weight of iron. The molecular weight of rust (Fe2O3) is 159.69 g/mol and the molecular weight of iron (Fe) is 55.85 g/mol. So, the weight of the rust formed is: \( 83 \, g \times \frac{159.69 \, g/mol}{55.85\, g/mol} = 237.96 \, g \).
03

Calculate the final mass of the iron bar and rust

To find the final mass of the iron bar and the rust, subtract the weight of the iron that turned into rust from the initial weight of the bar and then add the weight of the rust formed. So, the final weight is: \( 664 \, g - 83 \, g + 237.96 \, g = 818.96 \, g \). Thus, the final mass of the iron bar and the rust is 818.96 g.

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Key Concepts

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

Stoichiometry
Stoichiometry is like the recipe for a chemical reaction, telling you how much of each substance you need to combine to create certain products. Just like in cooking, where you might need a cup of flour for every two eggs to make a cake, stoichiometry gives us the proportions of reactants and products involved in chemical reactions.

For our iron bar example, to find out how much rust is formed from iron, we use stoichiometry to calculate the amount of product (rust) based on the quantity of the reactant (iron) that actually reacted. It's a critical tool to predict the outcome of reactions and understand how elements come together to form new compounds.
Molecular Weight
Think of molecular weight like a fruit basket that has to be weighed to know how heavy it is. Every molecule has its own 'basket weight,' which we call the molecular weight. It represents the sum of the atomic weights of all the atoms in a molecule. It's typically measured in grams per mole (g/mol).

Knowing the molecular weight allows us to convert between grams and moles, crucial for calculating how much of each chemical you'd get or need in a reaction. In solving for the rust problem, we used molecular weights to convert iron into its equivalent rust weight, since they have different molecular 'basket weights.'
Chemical Reaction
A chemical reaction is much like a dance where partners exchange to form new pairs. In chemistry, these partners are atoms or molecules that undergo a transformation to become new substances with different properties. For the iron bar exposed to moist air, a chemical reaction with oxygen leads to rust formation, changing the shiny metal into a flaky orange substance. We track these changes using chemical equations which depict the reactants on the left side, the products on the right, and the transformation process in the middle.
Rust Formation
Rust formation is the chemical equivalent of metal fatigue, where iron's exposure to moisture and oxygen causes it to degrade. The chemical process, commonly known as rusting, involves a reaction between iron (Fe) and oxygen (O2) in the presence of water (H2O) or moisture in the air. Rust (iron(III) oxide, Fe2O3) is not as strong or shiny as iron, and its formation means that metal objects gradually waste away if they are not protected. In our example, we see that an eighth of the iron bar turned into rust, which significantly changes both the appearance and the mass of the original bar.

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Most popular questions from this chapter

How many atoms are there in 5.10 moles of sulfur (S)?

A certain metal oxide has the formula MO, where \(\mathrm{M}\) denotes the metal. A 39.46-g sample of the compound is strongly heated in an atmosphere of hydrogen to remove oxygen as water molecules. At the end, \(31.70 \mathrm{~g}\) of the metal is left over. If \(\mathrm{O}\) has an atomic mass of 16.00 amu, calculate the atomic mass of \(\mathrm{M}\) and identify the element.

Pheromones are a special type of compound secreted by the females of many insect species to attract the males for mating. One pheromone has the molecular formula \(\mathrm{C}_{19} \mathrm{H}_{38} \mathrm{O}\). Normally, the amount of this pheromone secreted by a female insect is about \(1.0 \times 10^{-12} \mathrm{~g} .\) How many molecules are there in this quantity?

Tin(II) fluoride \(\left(\mathrm{SnF}_{2}\right)\) is often added to toothpaste as an ingredient to prevent tooth decay. What is the mass of \(\mathrm{F}\) in grams in \(24.6 \mathrm{~g}\) of the compound?

The fertilizer ammonium sulfate \(\left[\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\right]\) is prepared by the reaction between ammonia \(\left(\mathrm{NH}_{3}\right)\) and sulfuric acid: $$ 2 \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q) $$ How many kilograms of \(\mathrm{NH}_{3}\) are needed to produce \(1.00 \times 10^{5} \mathrm{~kg}\) of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} ?\)
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