C57H104O6(l)+O2(g) CO2(g)+H20(g) THE COEFFICIENTS OF PEANUT COMBUSTION
To determine the coefficients of the balanced equation for the combustion of peanut, we need to follow a systematic approach called balancing the equation. This involves adjusting the coefficients (numbers in front of the reactants and products) to ensure that the number of atoms of each element is the same on both sides of the equation.
Let's start by counting the number of atoms for each element on both sides of the equation:
Reactants:
C: 57
H: 104
O: 6 (from C57H104O6) + 2 (from O2) = 8
Products:
C: 1 (from CO2)
H: 2 (from H2O)
O: 3 (from CO2) + 1 (from H2O) = 4
Now, we need to balance the equation by adjusting the coefficients:
C57H104O6(l) + O2(g) → 57CO2(g) + 52H2O(g)
By multiplying the coefficients by the appropriate numbers, we ensure that the number of atoms of each element is balanced:
Reactants:
C: 57
H: 104
O: 6 (from C57H104O6) + 77 (from O2) = 83
Products:
C: 57 (from CO2)
H: 104 (from H2O)
O: 114 (from CO2) + 26 (from H2O) = 140
Now, the equation is balanced with the coefficients:
C57H104O6(l) + 83O2(g) → 57CO2(g) + 52H2O(g)
Therefore, the coefficients of the peanut combustion equation are C57H104O6 + 83O2 → 57CO2 + 52H2O.
You omitted the arrow which would cost you the problem in my class.
C57H104O6 + 80O2 ==> 57CO2 + 52H2O