-4 AI (s) + 302(g)=2Al2O3(s)How many grams of Al2O3 can be formed from 43.91 g of Al and 54.13 g of O2?(Hint: You need to determine which one is the limiting reactant).

According to the balanced equation, 108g of Al (4 x 27g) needs 96g of O2 (3 x 32g), so with a mathematical rule of three we can calculate the amount of oxygen that is needed to react with 43.91g of Al:

[tex]\begin{gathered} 108\text{gAl}-96gO_2 \\ 43.91\text{gAl}-x=\frac{43.91\text{gAl}\cdot96gO_2}{108\text{gAl}} \\ x=39.03gO_2 \end{gathered}[/tex]

Now we know that, 43.91g of Al will need 39.03g of O2 to react, and we have 54.13g of O2, so there is excess of oxygen.

Calculation with O2:

According to the balanced equation, 108g of Al (4 x 27g) needs 96g of O2 (3 x 32g), so with a mathematical rule of three we can calculate the amount of aluminum that is needed to react with 54.13g of O2:

[tex]\begin{gathered} 96gO_2-108gAl \\ 54.13gO_2-x=\frac{54.13gO_2\cdot108gAl}{96gO_2} \\ x=60.89\text{gAl} \end{gathered}[/tex]

Now we know that 54.13g of O2 will need 60.89g of Al to react, and we have only 43.91g of Al, so it is not sufficient, and that's why Al is the limiting reactant.

2nd) Now that we know that Al is the limiting reactant, we can calculate the grams of Al2O3 that will be produced in the reaction by using the limiting reactant (43.91g):

[tex]\begin{gathered} 108\text{gAl}-204gAl_2O_3 \\ 43.91\text{gAl}-x=\frac{43.91\text{gAl}\cdot204gAl_2O_3}{108\text{gAl}} \\ x=82.94gAl_2O_3 \end{gathered}[/tex]

In this calculation we use the balanced equation, so we know that 108g of Al will produce 204g of Al2O3 (2 x 102g).

Finally, 82.94g of Al2O3 will be produced from 43.91g of Al and 54.13g of O2.