Chemistry Reference
In-Depth Information
oxygen, it means that there is plenty of it. The reaction won't stop when we
run out of oxygen (because we won't!); it stops when we run out of pure
magnesium. In such a case, the reactant that will get completely used up is
called the limiting reactant, because it limits the amount of product that
will be produced.
Some of the examples that we will go over in this lesson will require
you to know how to balance and identify the types of chemical reactions
discussed in Chapter 6. Go back and review these reactions if you find that
you need to.
Example 1
How many grams of magnesium oxide (MgO) are formed when
2.00 g of magnesium (Mg) react with an excess of oxygen, according
to the balanced chemical reaction shown here?
2MgO + O
2
2MgO
Where do you begin? As in all examples of problem solving, start by
reading the question carefully and then taking the important information
out of the question. I recommend writing the appropriate information be-
low the equation, so that you don't have to keep referring back to the writ-
ten problem. The following shows the original problem with the important
information italicized. Notice how we should label the problem.
How many grams of magnesium oxide (MgO) are formed when 2.00 g
of magnesium (Mg) react with an excess of oxygen, according to the
balanced chemical reaction shown here?
2MgO
2.00 g excess x g
Because the problem says, “How many grams of magnesium oxide,” I
know that the magnesium oxide is the unknown, which I must solve for.
Therefore, I put “x g” under the magnesium oxide, indicating that I need to
find this mass. The oxygen, which is the excess reactant, plays no part in the
calculation. Therefore, I crossed it out, so that I will remember to ignore it.
I started with 2.00 grams of magnesium, my limiting reactant, so I put 2.00 g
under it to remind me of what I have.
Now, I know that the balanced chemical reaction is based on a molar
ratio, not a mass ratio, so I will need to change 2.00 grams of magnesium
into moles of magnesium using the calculation from Lesson 7-2. I divide
the mass of my sample by the molar mass of magnesium (24.3 g/mole),
which I can find on the periodic table.
2MgO + O
2
