Questions: Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients. CH3CH3(g) + O2(g) -> CO2(g) + H2O(g)

Solution Steps

The given chemical equation is: \[ \mathrm{CH}_{3} \mathrm{CH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \]

• Reactants:

  • Carbon (C): 2 (from \(\mathrm{CH}_{3} \mathrm{CH}_{3}\))
  • Hydrogen (H): 6 (from \(\mathrm{CH}_{3} \mathrm{CH}_{3}\))
  • Oxygen (O): 2 (from \(\mathrm{O}_{2}\))

• Products:

  • Carbon (C): 1 (from \(\mathrm{CO}_{2}\))
  • Hydrogen (H): 2 (from \(\mathrm{H}_{2} \mathrm{O}\))
  • Oxygen (O): 3 (1 from \(\mathrm{CO}_{2}\) and 2 from \(\mathrm{H}_{2} \mathrm{O}\))

To balance the carbon atoms, place a coefficient of 2 in front of \(\mathrm{CO}_{2}\): \[ \mathrm{CH}_{3} \mathrm{CH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \]

To balance the hydrogen atoms, place a coefficient of 3 in front of \(\mathrm{H}_{2} \mathrm{O}\): \[ \mathrm{CH}_{3} \mathrm{CH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{~g})+3\mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \]

Now, count the oxygen atoms in the products:

• Oxygen in \(\mathrm{CO}_{2}\): \(2 \times 2 = 4\)
• Oxygen in \(\mathrm{H}_{2} \mathrm{O}\): \(3 \times 1 = 3\)
• Total oxygen in products: \(4 + 3 = 7\)

To balance the oxygen atoms, place a coefficient of 3.5 in front of \(\mathrm{O}_{2}\): \[ \mathrm{CH}_{3} \mathrm{CH}_{3}(\mathrm{~g})+3.5\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{~g})+3\mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \]

Multiply all coefficients by 2 to eliminate the fraction: \[ 2\mathrm{CH}_{3} \mathrm{CH}_{3}(\mathrm{~g})+7\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4\mathrm{CO}_{2}(\mathrm{~g})+6\mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \]

Final Answer