Balance the given chemical reaction by short-cut method:
C6H6+O2=CO2+H2O.
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Dunno what you mean by short cut method, but I'll try to answer it the easiest I can.
Firstly, on left side we see C6H6, so there is 6C, but in right side, only one C, in CO2, so we put co efficient 6 for CO2. Similarly, we put 3 co efficient for H2O due to 6 H in left side.
Now, total oxygen on left side is 12 from 6CO2 and 3 from 3H20, which is a total of 15,so we give co efficient 15/2, on O2 in left side.
More precisely:
Element Reactants Product
C - - - - - 6 - - - - - 1
O - - - - - 2 - - - - - 3
H - - - - - 6 - - - - - 2
Based on this tabular column, if you orderly start from balancing C, H and finally O, as stated above, the reaction will be balanced as:
C6H6 + 7.5O2 - - > 6CO2 + 3H2O
If we multiply all co efficient by 2, it will remove decimal co efficients, but uses two moles of benzene.
Firstly, on left side we see C6H6, so there is 6C, but in right side, only one C, in CO2, so we put co efficient 6 for CO2. Similarly, we put 3 co efficient for H2O due to 6 H in left side.
Now, total oxygen on left side is 12 from 6CO2 and 3 from 3H20, which is a total of 15,so we give co efficient 15/2, on O2 in left side.
More precisely:
Element Reactants Product
C - - - - - 6 - - - - - 1
O - - - - - 2 - - - - - 3
H - - - - - 6 - - - - - 2
Based on this tabular column, if you orderly start from balancing C, H and finally O, as stated above, the reaction will be balanced as:
C6H6 + 7.5O2 - - > 6CO2 + 3H2O
If we multiply all co efficient by 2, it will remove decimal co efficients, but uses two moles of benzene.
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