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Aluminum & Iron(III) Oxide Reaction: Welding Heat & Calculations
Chapter 3, Problem 3.15(choose chapter or problem)
The reaction between aluminum and iron(III) oxide can generate temperatures approaching \(3000^{\circ} \mathrm{C}\) and is used in welding metals:
\(2 \mathrm{Al}+\mathrm{Fe}_{2} \mathrm{O}_{3} \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}+2 \mathrm{Fe}\)
In one process, \(124 \mathrm{~g}\) of \(\mathrm{Al}\) are reacted with \(601 \mathrm{~g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\).
(a) Calculate the mass (in grams) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) formed.
(b) How much of the excess reagent is left at the end of the reaction?
Questions & Answers
QUESTION:
The reaction between aluminum and iron(III) oxide can generate temperatures approaching \(3000^{\circ} \mathrm{C}\) and is used in welding metals:
\(2 \mathrm{Al}+\mathrm{Fe}_{2} \mathrm{O}_{3} \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}+2 \mathrm{Fe}\)
In one process, \(124 \mathrm{~g}\) of \(\mathrm{Al}\) are reacted with \(601 \mathrm{~g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\).
(a) Calculate the mass (in grams) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) formed.
(b) How much of the excess reagent is left at the end of the reaction?
ANSWER:Step 1 of 3
Given:
Mass of \(\mathrm{Al}=124 \mathrm{~g}\)
Mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}=601 \mathrm{~g}\)
Given reaction:
\(2 \mathrm{Al}+\mathrm{Fe}_{2} \mathrm{O}_{3} \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}+2 \mathrm{Fe}\)
(a) Here, we are asked to calculate the mass (in grams) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) formed:
Molar mass of \(\mathrm{Al}=26.981 \mathrm{~g} / \mathrm{mol}\).
Molar mass of \(\mathrm{Al}_{2} \mathrm{O}_{3}=101.96 \mathrm{~g} / \mathrm{mol} .\)
Molar mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}=159.69 \mathrm{~g} / \mathrm{mol}\).
We know the formula to calculate the number of moles:
Number of moles \((\mathrm{n})=\frac{\operatorname{mass}(\text { in } g)}{\text { Molarmass }}\)
and
Mass (in g) = Number of moles \(\times\) Molar mass
First, let's calculate the number of moles of each element:
The number of moles in \(Al\) is:
\(\begin{aligned}\mathrm{n} &=\frac{124 \mathrm{~g}}{26.981 \mathrm{~g} / \mathrm{mol}} \\ &=4.595 \text { moles }\end{aligned}\)
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Aluminum & Iron(III) Oxide Reaction: Welding Heat & Calculations
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Explore the chemical reaction between Aluminum and Iron (III) Oxide commonly used in welding. Learn to calculate the mass of Aluminum Oxide formed and identify the excess reagent. Dive deep into molar mass conversions stoichiometry and real-world applications in this chemistry tutorial.