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Fourth Example
The next example is more involved. It almost always seems that when you have a 3 or any
other odd number for a subscript it is going to give you some trouble.
Here we have magnesium nitride reacting with water to form magnesium oxide and ammonia.
Remember when you made magnesium oxide by burning magnesium in air? There was nitrogen in
the air, so some of the magnesium reacts with the nitrogen as well as with the oxygen. You
get magnesium oxide but you also get some magnesium nitride. By adding water to the
magnesium nitride and heating it, you can change it into magnesium oxide, and the nitrogen
goes away with the hydrogen as ammonia (NH3) which you can smell, if there is
enough.
First let’s take a quick inventory. We have three magnesiums on the
left and only one on the right, so that is going to take some work. Two nitrogens on the
left and one on the right; that is going to take some work. Two hydrogens on the left and
three on the right; so that is going to take some work. One oxygen on the left and one on
the right. Only the oxygen matches, and it may not stay that way when we start changing
everything else.
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Mg3N2 + H2O
 MgO +
NH3 |
| 3 Mg, 2 N, 2 H, 1 O |
1 Mg, 1 N, 3 H, 1 O |
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| There are two places where you could start balancing this equation. Notice
that you have two hydrogens on the left and three on the right. We could find the common
multiple and take three groups of two and two groups of three. This would work out
quickly. I will start with magnesium nitride, Mg3N2, because it is
the most complicated formula. With Mg3, we have three Mg on the left, so we
need three on the right; so put a three in front of the MgO.
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| Mg3N2 + H2O
3 MgO + NH3 |
| 3 Mg, 2 N, 2 H, 1 O |
3 Mg, 1 N, 3 H, 3 O |
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The three MgO also gives us three oxygens on the right. If we end up with
three oxygens on the right, we have to start with three on the left; so put a three in
front of the H2O. That gives us three oxygens but it also gives us six
hydrogens. If we start with six hydrogens on the left, we have to end up with six on the
right.
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| Mg3N2 + 3 H2O
3 MgO + NH3 |
| 3 Mg, 2 N, 6 H, 3 O |
3 Mg, 1 N, 3 H, 3 O |
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We have three in a group in the ammonia on the right, so we have to put a
two in front of the NH3. Two groups of three makes six hydrogens. It also gives
us two nitrogens. Looking back on the left, the formula is Mg3N2, so
we have two nitrogen on the left and two on the right. The balanced equation then is Mg3N2
+ 3 H2O
3 MgO + 2
NH3.
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| Mg3N2 + 3 H2O
3 MgO +
2 NH3 |
| 3 Mg, 2 N, 6 H, 3 O |
3 Mg, 2 N, 6 H, 3 O |
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To double check the answer we take another inventory. There are three magnesiums on the
left and three on the right. There are two nitrogens on the left and two on the right.
There are six hydrogens on each side. Finally oxygen, there are three oxygens from three
waters on the left and three MgO's give three oxygens on the right, so it is balanced.
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E-mail instructor:
Eden Francis
Clackamas Community College
©1998, 2002 Clackamas Community College, Hal Bender
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