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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

How is percentage yield calculated?

How many liters of \(\mathrm{O}_{2},\) density \(1.43 \mathrm{g} / \mathrm{L},\) are needed for the complete combustion of \(1.00 \mathrm{L} \mathrm{C}_{8} \mathrm{H}_{18},\) density 0.700 \(\mathrm{g} / \mathrm{mL} ?\)

Nitrogen monoxide, NO, reacts with oxygen to form nitrogen dioxide. Then the nitrogen dioxide reacts with oxygen to form nitrogen monoxide and ozone. Write the balanced equations. What is the theoretical yield in grams of ozone from 4.55 g of nitrogen monoxide with excess \(\mathrm{O}_{2} ?\) (Hint: First calcu- late the theoretical yield for NO \(_{2}\) , then use that value to calculate the yield for ozone.)

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