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Chapter 9: Problem 39

Iron and CO are made by heating 4.56 \(\mathrm{kg}\) of iron ore, \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) and carbon. The yield of iron is 88\(\% .\) How many kilograms of iron are made?

Short Answer

Expert verified
So, 2.80 kg of iron are produced.

Step by step solution

01

Calculate the moles of iron ore

To find out the moles of \(Fe2O3\), use the formula: Moles = Mass (kg) / Molar Mass First, convert the mass of \(Fe2O3\) in kg to g: 4.56 kg = 4560 g. Look up the molar mass of \(Fe2O3\) (it's 159.69 g/mol). Use the formula to find: Moles = 4560 g / 159.69 g/mol = 28.54 mol.
02

Calculate the mass of iron in the ore

Knowing that the molar mass of \(Fe\) is 55.85 g/mol and recognizing that there are 2 moles of \(Fe\) in each mole of \(Fe2O3\) , calculate the mass of iron in 4.56 kg of the iron ore: Iron mass = moles of \(Fe2O3\) x (2 moles \(Fe\) / 1 mole \(Fe2O3\)) x (55.85 g \(Fe\) / 1 mole \(Fe\)) = 3183.44 g = 3.18344 kg.
03

Calculate the actual mass of iron produced

The mass of iron actually obtained is given by the yield. So multiply the mass of the iron by the yield (88%): Iron obtained = Iron mass x Yield/100 = 3.18344 kg x 88/100 = 2.80142 kg.

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Key Concepts

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

Understanding the Mole Concept
The mole concept is a fundamental principle in chemistry that relates the mass of substances to the number of particles they contain. It allows chemists to

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Most popular questions from this chapter

Nitrogen dioxide from exhaust reacts with oxygen to form ozone. What mass of ozone could be formed from 4.55 \(\mathrm{g} \mathrm{NO}_{2} ?\) If only 4.58 \(\mathrm{g} \mathrm{O}_{3}\) formed, what is the percentage yield? $$\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{O}_{3}(g)$$

Why is the term limiting used to describe the limiting reactant?

Use the equation provided to answer the questions that follow. The density of oxygen gas is 1.428 \(\mathrm{g} / \mathrm{L}\) . $$2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$$ \begin{equation}\begin{array}{l}{\text { a. What volume of oxygen can be made }} \\ {\text { from } 5.00 \times 10^{-2} \text { mol of } \mathrm{KClO}_{3} ?} \\ {\text { b. How many grams } \mathrm{KClO}_{3} \text { must react to }} \\\ {\text { form } 42.0 \mathrm{mL} \mathrm{O}_{2} \text { ? }} \\ {\text { c. How many milliliters of } \mathrm{O}_{2} \text { will form at }} \\ {\text { STP from } 55.2 \mathrm{g} \mathrm{KClO}_{3} ?}\end{array}\end{equation}

Write the conversion factor needed to convert from g \(\mathrm{O}_{2}\) to \(\mathrm{L} \mathrm{O}_{2}\) if the density of \(\mathrm{O}_{2}\) is 1.429 \(\mathrm{g} / \mathrm{L}\) .

The following reaction can be used to remove \(\mathrm{CO}_{2}\) breathed out by astronauts in a spacecraft. $$2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)$$ \begin{equation}\begin{array}{l}{\text { a. How many grams of carbon dioxide can }} \\ {\text { be removed by } 5.5 \text { mol LiOH? }} \\ {\text { b. How many milliliters } \mathrm{H}_{2} \mathrm{O} \text { (density }=} \\\ {0.997 \mathrm{g} / \mathrm{mL} \text { ) could form from } 25.7 \mathrm{g} \text { LiOH? }} \\ {\text { c. How many molecules } \mathrm{H}_{2} \mathrm{O} \text { could be }} \\ {\text { made when } 3.28 \mathrm{g} \mathrm{CO}_{2} \text { react? }}\end{array}\end{equation}
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