Bancing chemical equation

Method 1 :
Step 1 :
Assign a "1" to C2H6.
1C2H6 +O2 → CO2 + H2O

Step 2 :
To balance C, assign a "2" to CO2 such that there are 2 Con both sides. To balance H, assign a "3" to H2O such thatthere are 6 H on both sides.
1C2H6 +O2 → 2CO2+ 3H2O

Step 3 :
To balance O, assign a "3.5" to O2 such that there are 7 Oon both sides.
1C2H6 +3.5O2 → 2CO2+ 3H2O

Step 4 :
To avoid that decimal 3.5, multiply all coefficients by 2.
2C2H6 +7O2 → 4CO2+ 6H2O


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Method 2 :
Step 1 :
By inspection, there are 6 H on the right hand side and thus the coefficient ofH2O is "3". Obviously, this leads to an odd number (3) ofO atom. However, the number of O must be even because of the O2 onthe left hand side. To avoid this, assign a "2" to the C2H6.
2C2H6 +O2 → CO2 + H2O

Step 2 :
To balance C, assign a "4" to CO2 such that there are 4 Con both sides. To balance H, assign a "6" to H2O such thatthere are 12 H on both sides.
2C2H6 +O2 → 4CO2+ 6H2O

Step 3 :
To balance O, assign a "7" to O2 such that there are 14 Oon both sides.
2C2H6 +7O2 → 4CO2+ 6H2O


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Method 3 :
Step 1 :
Apply the chemical equation: CxHy + (x + 0.25y)O2→ xCO2 + 0.5yH2O
Now, x = 2 and y = 6
Therefore, x + 0.25y = 3.5 and 0.5y = 3, i.e.
C2H6 + 3.5O2 → 2CO2+ 3yH2O

Step 2 :
To avoid that decimal 3.5, multiply all coefficients by 2.
2C2H6 +7O2 → 4CO2+ 6H2O

C2H6+ (3 1/2) O2 --> 2CO2 + 3H2O

因右手邊最少要有2個C,及6個H, 所以

平衡了這兩個之後, 左邊/右邊欠多少O

加上就是了