Question

Consider the following chemical reaction:

4 Al(s) + 3 O2(g) ⟶2 Al2O3(s)

If 4.2 grams of Al reacts with 3.0 grams of O2,

1. How many grams of Al2O3 could be produced?

2. Which reactant is limiting?

3. Which reactant is excess?

4. How much of the excess reactant would remain?

Answer 1

1. mAl₂O₃ = 6.42 g

2. O₂

3. Al.

4. mAl = 1.67 g

Step-by-step explanation:

Let's calculate the moles of Al and O2 first before doing any other calculations. The molecular weights of Al and O are 26.98 g/mol and 16 g/mol, so the moles are:

moles = mass / Atomic weight or molar mass

moles Al = 4.2 / 26.98 = 0.156 moles

moles O2 = 3 / (2*16) = 0.094 moles

The molar mass of Al2O3 is:

MM = (2 * 26.98) + (3*16) = 101.96 g/mol

With these data, let's determine each part:

1, 2 and 3. Grams of Al2O3, limiting and excess reactant:

To do this, we need to know which is the limiting reactant. this reactant will tell us how many moles are produced of Al2O3 and then, with the previously calculated molar mass, we can determine its mass. Let's see again the overall reaction taking place:

4Al + 3O₂ ---------> 2Al₂O₃

According to this reaction, 4 moles of Al reacts with 3 moles of oxygen, so:

If: 4 moles Al ---------> 3 moles O₂

Then: 0.156 moles ----------> X

X = 0.156 * 3/4 = 0.117 moles of O₂ required.

However, we do not have 0.117 moles of O₂, we have 0.094 moles, therefore we can conclude that the oxygen is the limiting reactant and Aluminum is the excess reactant.

Now, the moles produced of Al₂O₃ will be:

3 moles O₂ -------> 2 moles Al₂O₃

0.094 moles -------> Y

Y = 0.094 * 2/3 = 0.063 moles of Al₂O₃

Finally the mass would be:

mAl₂O₃ = 0.063 * 101.96

mAl₂O₃ = 6.42 g

4. mass of excess reactant:

The mass of excess reactant would be:

moles Al = 0.156 - 0.094 = 0.062 moles

mAl = 0.062 * 26.98

mAl = 1.67 g

Hope this helps