Aluminum reacts with oxygen to form aluminum oxide. Select the correct unbalanced skeleton equation for this reaction. A. Al(s) + 0(g) - AIO(s) B. Al203(s) - Al(s) + 02(8) C. Al(s) - 02(8) + Al203(s) D. Al(s) + 02(8) - Al203(s)

D. Al(s) + O₂(g) → Al₂O₃(s)

Explanation:

Aluminum is a solid metal, so it is written as Al(s).

Oxygen is a diatomic gas, so we write this compound as O₂(g).

Aluminum oxide has the formula Al₂O₃ because in oxides the oxidation number of oxygen atom is -2 and for aluminum, the oxidation number is 3. Thus, we write this compound as Al₂O₃(s).

Now, we have to found the chemical equation in which the reactants (left side) are Al(s) and O₂(g) while the product (right side) is Al₂O₃(s). From the options, we can see that the correct is (D):

Al(s) + O₂(g) → Al₂O₃(s)

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.