Write a balanced chemical equation for the combustion of trioleylglycerol. molecular formula C57H110O6
2C57H110O6 + 163O2 ==> 114CO2 + 110H2O
To balance a chemical equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Here's an explanation of how you can balance the combustion reaction of trioleylglycerol (C57H110O6):
Step 1: Write the unbalanced equation:
C57H110O6 + O2 → CO2 + H2O
Step 2: Count the number of atoms on each side of the equation to determine which elements need balancing.
Left Side (Reactants):
Carbon (C): 57
Hydrogen (H): 110
Oxygen (O): 6
Right Side (Products):
Carbon (C): 1
Hydrogen (H): 2
Oxygen (O): 2
Step 3: Begin balancing elements one at a time, starting with the least abundant. In this case, we'll start with Oxygen (O):
Left Side (Reactants):
Oxygen (O): 6
Right Side (Products):
Oxygen (O): 2
Since there are more oxygen atoms on the left side than the right side, we need to add a coefficient in front of the oxygen molecule on the right side to balance the equation. In this case, the least common multiple of 6 and 2 is 6. Therefore, we need to multiply the oxygen molecule on the right side by three.
C57H110O6 + O2 → 3CO2 + H2O
Now, let's check if the equation is balanced:
Left Side (Reactants):
Carbon (C): 57
Hydrogen (H): 110
Oxygen (O): 6
Right Side (Products):
Carbon (C): 3 x 1 = 3
Hydrogen (H): 2
Oxygen (O): 3 x 2 = 6
The carbon and hydrogen are not balanced yet, so let's balance those next:
Left Side (Reactants):
Carbon (C): 57
Right Side (Products):
Carbon (C): 3
We can balance the carbon atoms by multiplying the carbon compound on the right side by 19, and the hydrogen atoms by multiplying the water molecule on the right side by 19. This gives us:
19C57H110O6 + O2 → 57CO2 + 38H2O
Now, let's check if the equation is balanced one last time:
Left Side (Reactants):
Carbon (C): 57
Hydrogen (H): 110
Oxygen (O): 6
Right Side (Products):
Carbon (C): 57
Hydrogen (H): 38 x 2 = 76
Oxygen (O): 57 x 2 + 6 = 120
The equation is now balanced. Therefore, the balanced chemical equation for the combustion of trioleylglycerol (C57H110O6) is:
19C57H110O6 + 57O2 → 57CO2 + 38H2O