To balance the combustion reaction of propane (C₃H₈) with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O), we can follow these steps:
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Write the unbalanced equation: \[ C_3H_8 + O_2 \rightarrow CO_2 + H_2O \]
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Count the number of atoms of each element in reactants and products:
- Reactants: 3 Carbon (C), 8 Hydrogen (H), and an unknown amount of Oxygen (O).
- Products: The number of Carbon (C) and Water (H₂O) will determine how many Oxygens are needed.
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Balance Carbon (C): We have 3 Carbons in C₃H₈, so we will need 3 CO₂ to balance the Carbon atoms: \[ C_3H_8 + O_2 \rightarrow 3 CO_2 + H_2O \]
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Balance Hydrogen (H): We have 8 Hydrogens in C₃H₈, which means we need 4 H₂O to balance the Hydrogen atoms: \[ C_3H_8 + O_2 \rightarrow 3 CO_2 + 4 H_2O \]
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Now, count the Oxygen atoms on the right side:
- From 3 CO₂, we get \(3 \times 2 = 6\) Oxygen atoms.
- From 4 H₂O, we get \(4 \times 1 = 4\) Oxygen atoms.
- Total Oxygen on the right = \(6 + 4 = 10\) Oxygen atoms.
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To get 10 Oxygen atoms from O₂, since each O₂ molecule has 2 Oxygen atoms, this means we will need: \[ \frac{10}{2} = 5 \text{ O}_2 \]
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Now we have the complete balanced equation: \[ C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \]
Thus, the coefficients are:
- C₃H₈: 1
- O₂: 5
- CO₂: 3
- H₂O: 4
The final balanced equation is: \[ \boxed{1} C_3H_8 + \boxed{5} O_2 \rightarrow \boxed{3} CO_2 + \boxed{4} H_2O \]