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How many grams of aluminum burned if \[200.0{\text{ }}g\] of aluminum oxide formed?

Last updated date: 21st Jul 2024
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Hint: If we type the balanced equation between al giving \[A{l_2}{O_3}\],
The quantity of aluminum atoms is two and the quantity of oxygen atoms is three. As we recognize the atomic masses of aluminum as \[26.9815386{\text{ }}grams\] per mole and oxygen as \[16.9994{\text{ }}grams\] per mole.

Complete step by step answer:
Answer type 1st –
Balanced equation
\[4Al\left( s \right){\text{ }} + {\text{ }}3{O_2}\left( g \right)\xrightarrow{\Delta }2A{l_2}{O_3}\]
There are three steps compulsory to answer this question.
1. Define moles of \[A{l_2}{O_3}\] by dividing the specified mass by its molar mass \[\left( {101.960{\text{ }}g/mol} \right).\]I select to do this by means of multiplying by the reciprocal of the molar mass (mol/g).
2. Define moles of \[\;Al\;\] by multiplying moles \[A{l_2}{O_3}\] by the mole ratio between \[\;Al\;\]and \[A{l_2}{O_3}\] as of the balanced equation, with \[\;Al\;\] in the numerator.
3.Define mass of \[\;Al\;\] by multiplying moles \[\;Al\;\] by this one molar mass \[\left( {26.982g/mol} \right).\]

\[1.962mol{\text{ }}A{l_2}{O_3} \times \dfrac{{4molAl}}{{2molA{l_2}{O_3}}} = 3.924{\text{ }}mol{\text{ }}Al\]
\[3.924mol{\text{ }}Al \times \dfrac{{26.982g{\text{ }}Al}}{{1mol{\text{ }}Al}} = 105.9{\text{ }}g\](round to four significant figures)
We can association all steps into one equation:
\[200.0g{\text{ }}A{l_2}{O_3} \times \dfrac{{1mol{\text{ }}A{l_2}{O_3}}}{{101.960g{\text{ }}A{l_2}{O_3}}} \times \dfrac{{4mol{\text{ }}Al}}{{2mol{\text{ }}A{l_2}{O_3}}} \times \dfrac{{26.982g{\text{ }}Al}}{{1mol{\text{ }}Al}} = 105.9{\text{ }}g\]
(round to four significant figures)
\[105.9{\text{ }}g\]\[\;Al\;\]were burned in excess\[\;{O_{2\;}}\] to produce \[200.0{\text{ }}g\] \[A{l_2}{O_3}\]

Answer type 2nd -The molecular mass of Aluminum Oxide is nearly \[102{\text{ }}grams\]per mole.
\[2 \times Al = 2 \times 27 = 54\]
\[3 \times O = 3 \times 16 = 48\]
\[54 + 48 = 102\]
$\dfrac{{200}}{{102}} = 1.96{\text{moles}}$
There are two atoms of Aluminum in one mole and every atom has a mass of nearly \[27{\text{ }}grams\] so Mass of \[Al{\text{ }} = \;1.96 \times 2 \times 27 = 106\;grams\]

Note: Do not be confused with the confrontations alumina and aluminum oxide. Mutually are the same. Alumia is the Communal name given to aluminum oxide. Molecular formula of aluminum oxide is \[A{l_2}{O_3}\].Generally Alumina is presented in nature in the form of bauxite. Bauxite is an ore that is mined as a topsoil in several tropical and subtropical regions. In \[1887\] a process exposed called the Bayer process. It is the primary procedure by which alumina can be taken out from bauxite.