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If 45.3 grams of aluminum react with an excess of oxygen, as shown in the balanced chemical equation below, how many grams of aluminum oxide can be formed? Please show all your work for the calculations for full credit.

4Al + 3O2 yields 2Al2O3

Respuesta :

Mole of al=mass/mr
=45.3/27=1.6777
Mole of al2o3=2/4* moles of al
1.67777*2=mass/mr
Mass of al2o3=1.67777*2*(27*2+16*3)=342 g
Ondinne

Answer:

86g of [tex]Al_{2}O_{3}[/tex]

Explanation:

1. The balanced chemical equation is:

[tex]_{4}Al+_{3}O_{2}=_{2}Al_{2}O_{3}[/tex]

2. Use the stoichiometry of the reaction to calculate how many moles of aluminum oxide can be formed.

If the oxygen is in excess it means that the aluminum is the limiting reagent and the calculations must be made using the mass of aluminum.

[tex]45.3gAl*\frac{1molAl}{27gAl}*\frac{2molesAl_{2}O_{3}}{4molesAl}=0.84molesAl_{2}O_{3}[/tex]

3. Use the molar mass of the [tex]Al_{2}O_{3}[/tex] to calculate how many grams of aluminum oxide can be formed:

[tex]0.84molesAl_{2}O_{3}*\frac{102gAl_{2}O_{3}}{1molAl_{2}O_{3}}=86gAl_{2}O_{3}[/tex]