What mass of aluminum oxide can be recovered from the complete reaction of 31.5 g of Al with 31.5 g of oxygen gas? 4 Al (s) + 3 O2(g) → 2 Al2O3(s)

The answer to your question is  59.5 g

Explanation:

Chemical reaction

                             4Al   +   3O₂    ⇒     2Al₂O₃

Data

mass of Al₂O₃ = ?

mass of Al = 31.5 g

mass of O₂ = 31.5 g

Atomic mass Al = 4 x 27 = 108 g

Atomic mass O₂ = 6 x 16 = 96 g

Process

1.- Calculate the limiting reactant

Theoretical proportion   Al/O₂  = 108/96 = 1.125

Experimental proportion Al/O₂ = 31.5/31.5 = 1

The limiting reactant is Al because the proportion diminished

2.- Calculate the mass of Al₂O₃

molecular mass Al₂O₃ = (27 x 2) + (16 x 3) = 2(102 g) = 204 g

                   108 g  of Al 204 g of Al₂O₃

                   31.5 g of Al    x

                    x = (31.5 x 204)/108

                    x = 59.5 g of Al₂O₃  

the correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots

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.