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Originally Posted by Dancer4life26
Here's the problem that I need help with:
Ammonia can be prepared by the reaction of magnesium nitride with water. The products are ammonia and magnesium hydroxide.
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Mg3N2 + H2O --> Mg(OH)2 + NH3
So, you have have 3 Mg on left side, So put coefficient in front of the Mg on the right side to equal that:
Mg3N2 + H2O --> 3Mg(OH)2 + NH3
Now, you have 2 nitrogen on the left side and 1 on the right, so you need to make the one on the right side 2 by adding the coefficient to it.
Mg3N2 + H2O --> 3Mg(OH)2 + 2NH3
Now you have 2 H on the left side, but 12 on the right ( (3x2) + (2 x 3) = 12). So, add a coefficient of 6 to the water on the left side (6x2=12)
Mg3N2 + 6H2O --> 3Mg(OH)2 + 2NH3
Count the number of oxygens on each side to be sure they balance:
(6x1=6)(left side) (3x2=6) right side
therefore the balanced equation is:
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| Mg3N2 + 6H2O --> 3Mg(OH)2 + 2NH3 |
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-When trying to balance an equation, do the atoms other than H and O's first, as they are harder to balance, then deal with the oxygens and hydrogens...
And follow the same logic i used hereto do other similar problems