Courses
Courses for Kids
Free study material
Offline Centres
More

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

Last updated date: 21st Feb 2024
Total views: 337.5k
Views today: 10.37k
Verified
337.5k+ views
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.

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.