Question
hi!i am a f4 student in hk.i have know the basic concept of balance equation.but some too complex equation i cant balance.is there any general short cut for balancing any simple or complex chemical equation?
Answers
Steve
just look for least common multiples of the atom counts
For example, if you want to make a hydrogen balloon, a good reaction is lye and aluminum (try it!)
Al + NaOH = Al(NaO)<sub>3</sub> + H<sub>2</sub>
The problem is that the right side needs
H<sub>2</sub>
(NaO)<sub>3</sub>
Since you're mixing 2's and 3's, you will want to make multiples so that things get used up in 6's
2Al + 3NaOH = 2Al(NaO)<sub>3</sub> + 3H<sub>2</sub>
With more complex reactions, things do get more involved, but usually the numbers are relatively small, even with things like burning gasoline (octane):
2C<sub>8</sub>H<sub>18</sub>(g)+25O<sub>2</sub>(g) = 16CO<sub>2</sub>(g)+18H<sub>2</sub>O(g)
For example, if you want to make a hydrogen balloon, a good reaction is lye and aluminum (try it!)
Al + NaOH = Al(NaO)<sub>3</sub> + H<sub>2</sub>
The problem is that the right side needs
H<sub>2</sub>
(NaO)<sub>3</sub>
Since you're mixing 2's and 3's, you will want to make multiples so that things get used up in 6's
2Al + 3NaOH = 2Al(NaO)<sub>3</sub> + 3H<sub>2</sub>
With more complex reactions, things do get more involved, but usually the numbers are relatively small, even with things like burning gasoline (octane):
2C<sub>8</sub>H<sub>18</sub>(g)+25O<sub>2</sub>(g) = 16CO<sub>2</sub>(g)+18H<sub>2</sub>O(g)
kate
so what side should i look for the LCM?
kate
how about HNO3 +Cu ------->Cu(NO3)2+NO+H2O,PLEASE SHOW ME STEPS.
Steve
Note that Cu(NO3)2 requires 2 NO3 groups. So, the first step would be to make
2HNO3 +Cu ------->Cu(NO3)2+NO+H2O
Now you have the (NO3)2 taken care of.
On the right we still have NO+H2O = H2NO2, so we need another H2NO3 on the left:
3HNO3 +Cu ------->Cu(NO3)2+NO+H2O
Now we are left with an extra O on the left. So, for every 3 NO3 on the left we get 8 O on the right. Sounds like 24 is a good LCM
8 NO3 + Cu ----> Cu(NO3)2 + NO + 4H2O
We've used up all the H, but now need to fill in for the NO3 stuff.
3Cu+ 8HNO3----> 3Cu(NO3)2 + 2NO +4H2O
That was pretty trial-and-error. Maybe a web search on balancing reaction equations or stoichiometry will yields a more methodical method.
2HNO3 +Cu ------->Cu(NO3)2+NO+H2O
Now you have the (NO3)2 taken care of.
On the right we still have NO+H2O = H2NO2, so we need another H2NO3 on the left:
3HNO3 +Cu ------->Cu(NO3)2+NO+H2O
Now we are left with an extra O on the left. So, for every 3 NO3 on the left we get 8 O on the right. Sounds like 24 is a good LCM
8 NO3 + Cu ----> Cu(NO3)2 + NO + 4H2O
We've used up all the H, but now need to fill in for the NO3 stuff.
3Cu+ 8HNO3----> 3Cu(NO3)2 + 2NO +4H2O
That was pretty trial-and-error. Maybe a web search on balancing reaction equations or stoichiometry will yields a more methodical method.