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.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.