What is the chemical equation for the burning of propane if all the products and reactants are in the gas phase?

A. [tex]4 H_2 O(g) + 3 CO_2(g) + 5 O_2(g) \rightarrow C_3 H_8(g)[/tex]
B. [tex]C_3 H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2 O(g)[/tex]
C. [tex]C_3 H_8(g) + O_{10}(g) \rightarrow 3 CO_2(g) + 4 H_2 O(l)[/tex]
D. [tex]4 H_2 O(l) + 3 CO_2(g) \rightarrow C_3 H_8(g) + 5 O_2(g)[/tex]



Answer :

Certainly! To determine the correct chemical equation for the burning of propane ([tex]\(C_3H_8\)[/tex]), we need to identify which one matches the balanced combustion reaction of propane in the gas phase. The general form for the combustion of a hydrocarbon [tex]\(C_xH_y\)[/tex] in the presence of oxygen [tex]\(O_2\)[/tex] is:

[tex]\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \][/tex]

For propane, [tex]\(C_3H_8\)[/tex]:

[tex]\[ C_3H_8 + O_2 \rightarrow CO_2 + H_2O \][/tex]

Next, we balance the chemical equation to ensure we have the same number of each type of atom on both sides of the equation.

### Step-by-Step Balancing:

1. Write down the unbalanced equation:
[tex]\[ C_3H_8 + O_2 \rightarrow CO_2 + H_2O \][/tex]

2. Balance carbon (C) atoms:
- There are 3 carbon atoms in [tex]\(C_3H_8\)[/tex].
- On the product side, each [tex]\(CO_2\)[/tex] molecule has 1 carbon atom.
- Therefore, we need 3 [tex]\(CO_2\)[/tex] molecules to balance the carbon atoms.
[tex]\[ C_3H_8 + O_2 \rightarrow 3 CO_2 + H_2O \][/tex]

3. Balance hydrogen (H) atoms:
- There are 8 hydrogen atoms in [tex]\(C_3H_8\)[/tex].
- On the product side, each [tex]\(H_2O\)[/tex] molecule has 2 hydrogen atoms.
- Therefore, we need 4 [tex]\(H_2O\)[/tex] molecules to balance the hydrogen atoms.
[tex]\[ C_3H_8 + O_2 \rightarrow 3 CO_2 + 4 H_2O \][/tex]

4. Balance oxygen (O) atoms:
- On the product side, we now have:
- 3 [tex]\(CO_2\)[/tex] molecules, each contributing 2 oxygen atoms for a total of [tex]\(3 \times 2 = 6\)[/tex] oxygen atoms.
- 4 [tex]\(H_2O\)[/tex] molecules, each contributing 1 oxygen atom for a total of [tex]\(4 \times 1 = 4\)[/tex] oxygen atoms.
- Therefore, we need [tex]\(6 + 4 = 10\)[/tex] oxygen atoms in total.
- On the reactant side, [tex]\(O_2\)[/tex] is a diatomic molecule, so each [tex]\(O_2\)[/tex] contributes 2 oxygen atoms.
- To balance, we need [tex]\(10 \div 2 = 5\)[/tex] [tex]\(O_2\)[/tex] molecules.
[tex]\[ C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \][/tex]

### Verification:

1. Count each type of atom on both sides of the equation:
- Reactants: [tex]\(C_3H_8 + 5 O_2\)[/tex]
- Carbon: 3
- Hydrogen: 8
- Oxygen: [tex]\(5 \times 2 = 10\)[/tex]

- Products: [tex]\(3 CO_2 + 4 H_2O\)[/tex]
- Carbon: 3
- Hydrogen: 8
- Oxygen: [tex]\(3 \times 2 + 4 \times 1 = 6 + 4 = 10\)[/tex]

All atoms balance correctly on both sides of the equation, verifying the equation is correct.

### Conclusion:
The balanced chemical equation for the burning of propane where all reactants and products are in the gas phase is:

[tex]\[ \boxed{C_3H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(g)} \][/tex]