How many of grams of aluminum would be produced from the decomposition of 21.0 g of Al2O3
Here is a solved example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find out how many grams of aluminum would be produced from the decomposition of 21.0 g of Al2O3 (aluminum oxide), we need to use the balanced equation for the reaction.
The balanced equation for the decomposition of aluminum oxide is:
2 Al2O3 → 4 Al + 3 O2
From the balanced equation, we can see that 2 moles of Al2O3 will produce 4 moles of Al.
First, we need to calculate the number of moles of Al2O3 using its molar mass. The molar mass of Al2O3 is calculated by adding the atomic masses of aluminum (Al) and oxygen (O) together.
Al2O3:
Al (aluminum) atomic mass = 26.98 g/mol
O (oxygen) atomic mass = 16.00 g/mol
Molar mass of Al2O3 = (2 * Al atomic mass) + (3 * O atomic mass)
= (2 * 26.98 g/mol) + (3 * 16.00 g/mol)
= 52.96 g/mol + 48.00 g/mol
= 100.96 g/mol
Next, we can calculate the number of moles of Al2O3 by dividing its mass by its molar mass.
Number of moles of Al2O3 = Mass of Al2O3 / Molar mass of Al2O3
= 21.0 g / 100.96 g/mol
≈ 0.208 moles
Now that we know the number of moles of Al2O3, we can determine the number of moles of Al produced using the stoichiometry of the balanced equation.
According to the equation, 2 moles of Al2O3 produce 4 moles of Al. Therefore, the number of moles of Al produced will be:
Number of moles of Al = Number of moles of Al2O3 * (4 moles of Al / 2 moles of Al2O3)
= 0.208 moles * (4/2)
= 0.416 moles
Finally, we need to calculate the mass of Al produced from the number of moles of Al using its molar mass.
Molar mass of Al = 26.98 g/mol
Mass of Al = Number of moles of Al * Molar mass of Al
= 0.416 moles * 26.98 g/mol
≈ 11.29 grams
Therefore, approximately 11.29 grams of aluminum would be produced from the decomposition of 21.0 grams of Al2O3.