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.
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?
6 answers
I need the answer
7.35g of KClO3 will begin required
7.35g of KClO3 will be required
6.516g
6.1275g