Aluminium reacts with oxygen to form aluminium oxide.How many grams of potassium chlorate would be heated to produce enough oxygen to form 5.1 grams of aluminium oxide?
7.35g of KClO3 will be required
7.35g of KClO3 will begin required
4Al + 3O2 = 2Al2O3
2KClO3 --> 2KCl + 3O2
5.1g Al2O3 = 5.1/101.96 = 0.050 moles Al2O3
So, you need 3/2 * 0.050 = 0.075 moles of O2.
Thus, you need 2/3 * 0.075 = 0.050 moles KClO3 to produce that needed O2.
I need the answer
6.1275g
To find out how many grams of potassium chlorate would need to be heated to produce enough oxygen to form 5.1 grams of aluminum oxide, we need to follow the steps below:
1. Write the balanced equation for the reaction:
4 Al + 3 O2 → 2 Al2O3
2. Calculate the molar mass of aluminum oxide (Al2O3):
Atomic mass of Al = 26.98 g/mol
Atomic mass of O = 16.00 g/mol
Molar mass of Al2O3 = (2 * Atomic mass of Al) + (3 * Atomic mass of O)
= (2 * 26.98 g/mol) + (3 * 16.00 g/mol)
= 101.96 g/mol
3. Determine the molar mass of oxygen in the reaction:
Since each oxygen molecule (O2) has a molar mass of 32 g/mol, the molar mass of oxygen in the reaction is 32 g/mol.
4. Use stoichiometry to find the ratio between oxygen and aluminum oxide:
From the balanced equation, we know that 3 moles of O2 are needed to produce 2 moles of Al2O3.
5. Calculate the number of moles of aluminum oxide:
Moles of Al2O3 = Mass of Al2O3 / Molar mass of Al2O3
= 5.1 g / 101.96 g/mol
≈ 0.050 moles
6. Determine the number of moles of oxygen needed:
Since the ratio between oxygen and aluminum oxide is 3:2, we can calculate:
Moles of O2 = (3/2) * Moles of Al2O3
= (3/2) * 0.050 moles
= 0.075 moles
7. Convert moles of oxygen to grams:
Grams of O2 = Moles of O2 * Molar mass of O2
= 0.075 moles * 32 g/mol
= 2.4 grams
Therefore, you would need to heat 2.4 grams of potassium chlorate to produce enough oxygen to form 5.1 grams of aluminum oxide.