Place the coefficients into the underlined spaces to balance the chemical equation. Not all coefficients must be used and some may be used more than once.

C3H8 (8)+ CO2(g) + H2O (l)
O2 (g)
1
2
3
4
5

1 answer

To balance the chemical equation for the combustion of propane (C₃H₈), we need to ensure that the number of atoms for each element is equal on both sides of the equation.

The unbalanced equation is:

\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

  1. Start by balancing the carbon (C) atoms:

    • Propane has 3 carbon atoms, so we need 3 CO₂ molecules: \[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + \text{H}_2\text{O} \]
  2. Next, balance the hydrogen (H) atoms:

    • Propane has 8 hydrogen atoms, which means we need 4 H₂O molecules to balance the H: \[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]
  3. Now, count the oxygen (O) atoms needed on the product side:

    • From 3 CO₂, we have \(3 \times 2 = 6\) O.
    • From 4 H₂O, we have \(4 \times 1 = 4\) O.
    • Total O atoms needed = 6 + 4 = 10 O.
  4. Now, find the number of O₂ molecules needed:

    • Since O₂ has 2 oxygen atoms per molecule, we need \(10 \div 2 = 5\) O₂ molecules: \[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]

The balanced equation is:

\[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]

So, the coefficients placed in the underlined spaces are:

  • C₃H₈: 1
  • O₂: 5
  • CO₂: 3
  • H₂O: 4

Thus, the final answer with coefficients would be:

\[\underline{1}\text{ C}_3\text{H}_8 + \underline{5}\text{ O}_2 \rightarrow \underline{3}\text{ CO}_2 + \underline{4}\text{ H}_2\text{O}\]