98 g
Explanation:
Start with the balanced equation. Then, make a little chart under the equation showing the information you have and need.
    3O₂(g) + 4Al(s)  →  2Al₂O₃(s)
m     ?      110 g
M Â Â Â ___ Â Â Â ____
n    ___ <=  ____
"m" is for mass. "M" is for molar mass (some teachers use "MM"). "n" is for the number of moles.
To find the mass of oxygen:
Calculate the molar mass of oxygen (
)Calculate molar mass of aluminum (
)Use
to find the moles of aluminum (
)With
, use the mole ratio to find the moles of oxygen (
)Use
and
to find the mass of oxygen (
)
To find molar mass, use the atomic mass on your periodic table. For each atom of an element, add on its atomic mass.
Molar mass of aluminum (one Al atom):
Molar mass of oxygen (two O atoms):
Update the chart:
        3O₂(g)    +      4Al(s)      →   2Al₂O₃(s)
m         ?             110 g
M    32.000 g/mol    26.982 g/mol
n         ___     <=    ____
Find the moles of aluminum
  Multiply mass by molar mass to find moles.
   The units "g" cancel out.
    Keep one extra significant figure. (110 has 2 sig. figs.)
        3O₂(g)    +      4Al(s)      →   2Al₂O₃(s)
m         ?             110 g
M    32.000 g/mol    26.982 g/mol
n         ___     <=  4.0(7) mol
Find the moles of oxygen using the mole ratio, which comes from the coefficients in the balanced equation.
The mole ratio of oxygen to aluminum is 3 to 4.
  Multiply moles of aluminum by the mole ratio.
      "molAl" units cancel out.
       Keep two sig. figs. when the first is a "5"
        3O₂(g)    +      4Al(s)      →   2Al₂O₃(s)
m         ?             110 g
M    32.000 g/mol    26.982 g/mol
n      3.0(52) mol  <=   4.0(7) mol
Find the mass of oxygen
    Multiply moles by molar mass.
      The "mol" unit cancels out.
            Keep one sig. fig. to round. "6" rounds up.
               <= Final answer
∴ 98 grams of oxygen are required to completely react with 110 grams of aluminum.