How many grams of aluminum is produced from the decomposition of 34grams of aluminum oxide
First you must understand that this is not a proper question because Al2O3 is not decomposed by heat. However, this is how you work this hypothetical question.
2Al2O3 ==> 4Al + 3O2
mols Al2O3 = grams/molar mass = ?
mols Al produced = 2*mols Al2O3
g Al = mols Al x atomic mass Al = ?
To find out how many grams of aluminum are produced from the decomposition of a given amount of aluminum oxide, you will need to use the balanced chemical equation for the reaction.
The balanced chemical equation for the decomposition of aluminum oxide is:
2Al2O3 -> 4Al + 3O2
From this equation, we see that 2 moles of aluminum oxide decompose to form 4 moles of aluminum. The molar mass of aluminum oxide (Al2O3) is 101.96 g/mol.
To calculate the number of moles of aluminum oxide, divide the given mass by its molar mass:
34 g Al2O3 / (101.96 g/mol) = 0.333 mol Al2O3
Since there is a 2:4 mole ratio between aluminum oxide and aluminum, we can determine the number of moles of aluminum produced by multiplying the moles of aluminum oxide by the ratio:
0.333 mol Al2O3 x (4 mol Al / 2 mol Al2O3) = 0.666 mol Al
Finally, to find the mass of aluminum produced, multiply the moles of aluminum by its molar mass (27 g/mol):
0.666 mol Al x (27 g/mol) = 18 g Al
Therefore, from the decomposition of 34 grams of aluminum oxide, approximately 18 grams of aluminum are produced.