… C3H8(g) + … O2(g) → … H2O(g) + … CO2(g) When the equation for the reaction represented above is ba — Stoichiometry Chemistry Question

STEPS:

1. Identify the Reaction Type: The provided equation represents the complete combustion of a hydrocarbon (propane, $C_3H_8$). In this type of reaction, a hydrocarbon reacts with oxygen gas ($O_2$) to produce carbon dioxide ($CO_2$) and water ($H_2O$).
2. Start with the Most Complex Molecule: Begin by assuming a coefficient of 1 for propane ($C_3H_8$). This gives you 3 carbon atoms and 8 hydrogen atoms on the reactant side.
3. Balance Carbon Atoms: To balance the 3 carbon atoms from the propane, you must have 3 carbon atoms on the product side. Since each $CO_2$ molecule contains one carbon, place a coefficient of 3 in front of $CO_2$:
$$1\ C_3H_8 + \dots O_2 \rightarrow \dots H_2O + \mathbf{3}\ CO_2$$
4. Balance Hydrogen Atoms: There are 8 hydrogen atoms in the propane molecule. To balance these on the product side, where hydrogen is found in water ($H_2O$), you need a coefficient that results in 8 atoms total. Since each water molecule has 2 hydrogens, use a coefficient of 4 ($4 \times 2 = 8$):
$$1\ C_3H_8 + \dots O_2 \rightarrow \mathbf{4}\ H_2O + 3\ CO_2$$
5. Count the Oxygen Atoms on the Product Side: Now that the products are balanced relative to the propane, total the oxygen atoms present:
* From $4\ H_2O$: $4 \times 1 = \mathbf{4}$ oxygen atoms.
* From $3\ CO_2$: $3 \times 2 = \mathbf{6}$ oxygen atoms.
* Total oxygen atoms needed = $4 + 6 = \mathbf{10}$ oxygen atoms.
6. Balance Oxygen Atoms on the Reactant Side: Oxygen exists as a diatomic molecule ($O_2$) on the reactant side. To obtain the 10 atoms required, you need 5 molecules of $O_2$ ($5 \times 2 = 10$):
$$\mathbf{1}\ C_3H_8 + \mathbf{5}\ O_2 \rightarrow \mathbf{4}\ H_2O + \mathbf{3}\ CO_2$$
7. Final Verification: Check that all coefficients (1, 5, 4, 3) are in their lowest whole-number terms and that the atoms on both sides match (C: 3=3; H: 8=8; O: 10=10). The coefficient for $O_2$ is 5, matching option D.

WHY_OTHERS_WRONG:

• A) 1, B) 2, and C) 3: These coefficients provide 2, 4, and 6 oxygen atoms respectively. None of these amounts are sufficient to balance the 10 oxygen atoms required by the 3 molecules of $CO_2$ and 4 molecules of $H_2O$ produced by the combustion of one propane molecule.
• E) 6: This coefficient would provide 12 oxygen atoms on the reactant side, which exceeds the 10 atoms needed for a balanced equation using the lowest whole-number terms.