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# How many grams of aluminum are in $25.0{\text{ }}g$ of aluminum oxide?

Last updated date: 13th Jun 2024
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Answer
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Hint: Aluminum oxide has a chemical formula $A{l_2}{O_3}$ . That's $\;2$ aluminum atoms used for every $\;3$ oxygen atoms It is the utmost commonly occurring of some aluminium oxides, and definitely identified as aluminium $\left( {III} \right)$ oxide.

Complete step by step answer:
First you have to invent how many moles of $A{l_2}{O_3}$ are in $25.0{\text{ }}g$ . To find that you divide the quantity of grams by the molar mass.
molar mass $= {\text{ }}2\left( {26.98} \right){\text{ }} + {\text{ }}2\left( {16} \right){\text{ }} = {\text{ }}85.96$
At present to find the number of moles of $A{l_2}{O_3}$ you have...
$25{\text{ }}g{\text{ }}/{\text{ }}\left( {85.96g{\text{ }}/mol} \right){\text{ }} = {\text{ }}.291{\text{ }}moles$ (the grams cancelled on view)
Now that ratio of the quantity of Aluminum to $A{l_2}{O_3}$ is $2:1$ ( $2{\text{ }}moles$ of $\;Al$ for $1{\text{ }}mole$ of $A{l_2}{O_3}$ )
so you multiply the number of moles of $A{l_2}{O_3}$ by $\;2$ .
$.291{\text{ }}\left( 2 \right){\text{ }} = {\text{ }}.582{\text{ }}moles$ of $\;Al$
now Convert rear to grams by multiplying the number of moles of $\;Al$ by the molar mass of $\;Al$
So $.582{\text{ }}mol{\text{ }}x{\text{ }}26.98{\text{ }} = {\text{ }}15.69{\text{ }}g$

Additional information:
It's time that we take a profounder appearance into one of the inorganic compounds like aluminum oxide. It is similarly known as aluminum oxide and generally comes in the crystalline powdery form. It is insoluble to water, and together chemical and physical belongings vary based on the process of preparation. The crystalline modifications are based on the dissimilar methods of preparation. As all varieties are molded, very high temperatures are applied to turn out to be chemically inert.

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.