Answer
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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.
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.