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Chapter 3: Problem 154

A major industrial use of hydrochloric acid is in metal pickling. This process involves the removal of metal oxide layers from metal surfaces to prepare them for coating. (a) Write an equation between iron(III) oxide, which represents the rust layer over iron, and \(\mathrm{HCl}\) to form iron(III) chloride and water. (b) If 1.22 moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(289.2 \mathrm{~g}\) of HCl react, how many grams of \(\mathrm{FeCl}_{3}\) will be produced?

Short Answer

Expert verified
The balanced chemical equation is \(Fe_{2}O_{3} + 6HCl \rightarrow 2FeCl_{3} + 3H_{2}O\), and 396.2 g of \(FeCl_{3}\) will be produced from 1.22 moles of \(Fe_{2}O_{3}\) and 289.2 g of HCl.

Step by step solution

01

Write the Unbalanced Chemical Equation

First write the chemical equation for the reaction between iron(III) oxide (\(Fe_{2}O_{3}\)) and hydrochloric acid (\(HCl\)), producing iron(III) chloride (\(FeCl_{3}\)) and water (\(H_{2}O\)). This comes out as \(Fe_{2}O_{3} + HCl \rightarrow FeCl_{3} + H_{2}O\).
02

Balance the Chemical Equation

Next balance the equation. A balanced equation keeps the same number of atoms for each element on both sides of the equation. The balanced equation becomes \(Fe_{2}O_{3} + 6HCl \rightarrow 2FeCl_{3} + 3H_{2}O\).
03

Convert grams of HCl to moles

Before we start calculating the amount of \(FeCl_{3}\) that will be produced, we need to convert the grams of HCl into moles. The molar mass of \(HCl\) is approximately \(36.46 \, g/mol\). So we divide the given mass of \(HCl\) (289.2 g) by its molar mass to get the number of moles: \(289.2 \, g / 36.46 \, g/mol = 7.93 \, mol\).
04

Use the Balanced Chemical Equation to Determine the Amount of \(FeCl_{3}\) Produced.

Now we can use the balanced chemical equation to find out how many moles of \(FeCl_{3}\) would be produced by 1.22 moles of \(Fe_{2}O_{3}\) and 7.93 moles of \(HCl\). According to the balanced equation, 1 mole of \(Fe_{2}O_{3}\) reacts with 6 moles of \(HCl\) to produce 2 moles of \(FeCl_{3}\). Since the \(Fe_{2}O_{3}\) is the limiting reactant, it will determine the amount of \(FeCl_{3}\) produced: \(1.22\,moles \times 2 = 2.44\,moles \, of \, FeCl_{3}\).
05

Convert Moles of \(FeCl_{3}\) to Grams

Finally, convert the moles of \(FeCl_{3}\) into grams. The molar mass of \(FeCl_{3}\) is approximately \(162.2\,g/mol\). So we multiply the number of moles of \(FeCl_{3}\) by its molar mass to get the weight in grams: \(2.44\,moles \times 162.2\,g/mol = 396.2\,g\).

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