How many moles of oxygen are needed to react with 87 grams of aluminum

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.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )