Consider the reaction: Fe2O3+2Al=Al2O3
How many grams of Al are necessary to make 100.2 grams of Al2O3?
mols Al2O3 = ?
Convert mols Al2O3 to mols Al
Convert mols Al to grams Al. See your post above if you need refreshing. Same process. All of these stoichiometry problems are done the same way.
To determine the number of grams of Al required to produce 100.2 grams of Al2O3, you need to use stoichiometry. Stoichiometry is a method that relates the quantities of substances involved in a chemical reaction.
First, you need to balance the given chemical equation:
Fe2O3 + 2Al = Al2O3
The balanced equation shows that the ratio of Al to Al2O3 is 2:1, meaning that for every 2 moles of Al used, you'll obtain 1 mole of Al2O3.
To find the molar mass of Al2O3, you need to calculate the sum of the atomic masses of its constituents: Al (26.98 g/mol) and O (16.00 g/mol).
Molar mass of Al2O3 = (2 * Al) + (3 * O) = (2 * 26.98) + (3 * 16.00) = 101.96 g/mol.
Now, you can use stoichiometry to calculate the amount of Al required:
1 mole of Al2O3 weighs 101.96 g.
Therefore, 100.2 g of Al2O3 is equivalent to (100.2 g / 101.96 g/mol) = 0.982 moles of Al2O3.
Since the ratio of Al to Al2O3 is 2:1, you need half the number of moles of Al as moles of Al2O3. Therefore, you'll need 0.982 * 0.5 = 0.491 moles of Al.
Finally, to determine the mass of Al, you can multiply the number of moles by the molar mass of Al (26.98 g/mol):
Mass of Al = 0.491 moles * 26.98 g/mol = 13.24 grams.
Therefore, you would need approximately 13.24 grams of Al to produce 100.2 grams of Al2O3.