Answer:
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Explanation:
Chemical equation:
4Al(s) + 3O₂(l) → 2AlO₃(s)
Given data:
Mass of aluminium = 87 g
Moles of oxygen needed = ?
Solution:
Moles of aluminium:
Number of moles of aluminium= Mass/ molar mass
Number of moles of aluminium= 87 g/ 27 g/mol
Number of moles of aluminium= 3.2 mol
Now we will compare the moles of aluminium with oxygen.
Al : O₂
4 : 3
3.2 : 3/4×3.2 = 2.4 mol
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Answer:
I think the answer is 368.544 centigrams
Explanation:
I hope I helped
I know what you're asking but I don't think the question is stated properly. Technically, an atom will not join with an "oxide" ion; i.e., the oxide ion is an atom of oxygen to which two electrons have been added. An oxide ion will add to 2 K ions or 1 Ca ion. The K ion has lost just one electron so it takes two of them to equal the 2- charge on the oxide ion whereas the Ca ion has lost two electrons and it takes only one of them to equal the charge on the oxide ion.
Answer:
There are many properties that differentiate metals from non-metals. They are: Luster, conductivity of heat and electricity, physical strength i.e. Brittle or difficult to break, etc.
