Answer:
a. formaldehyde (CH₂O(g)): CH₂O(g) + O₂(g) → CO₂(g) + H₂O(g)
b. heptane (C₇H₁₇(l)): 4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)
c. benzene (C6H6(l)): C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3 H₂O(g)
Explanation:
In a reaction of combustion, a hydrocarbon compound (composed of C, H and O) reacts with oxygen gas (O₂). The complete combustion of a hydrocarbon - such as formaldehyde, heptane, and benzene - produces carbon dioxide (CO₂) and water (H₂O).
Thus, we write the reactants and products for each combustion reaction and then we balance the atoms: C, H, and O.
a. formaldehyde (CH₂O(g)):
CH₂O(g) + O₂(g) → CO₂(g) + H₂O(g)
The chemical equation is already balanced with the coefficients 1: we have the same number of C atoms (1), H (2) and O (3) on both sides of the equation.
b. heptane (C₇H₁₇(l)):
C₇H₁₇(l) + O₂(g) → CO₂(g) + H₂O(g)
Here we have to write a coefficient 28 for CO₂ to balance the C atoms in the products side, and for C₇H₁₇ we write 28/4 = 7. With similar reasoning we found the coefficients for O₂ and H₂O:
4C₇H₁₇(l) + 45O₂(g) → 28CO₂(g) + 34H₂O(g)
c. benzene (C6H6(l)):
C₆H₆(l) + O₂(g) → CO₂(g) + H₂O(g)
First, we write a coefficient of 6 for CO₂ to balance the C atoms. Then, we have to balance H atoms: we write a coefficient 3 in H₂O. Now, we have 12 + 3 = 15 atoms of O on the reactants side. So, we write a half of these number of atoms in the coefficient for O₂: 15/2. We obtain the balanced equation:
C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3 H₂O(g)
