Chapter 15.2, Problem 7PP

To determine

Amount of water formed when methane reacts with oxygen

Answer to Problem 7PP

  $54\text{ g}$

Explanation of Solution

Given:

Amount of methane is $24\text{ g}$ , or $1.5\text{ mole}$ .

Amount of carbon dioxide is $66\text{ g}$ , or $1.5\text{ mole}$ .

Amount of oxygen is $96\text{ g}$ , or $3\text{ moles}$ .

Formula used:

The chemical equation when methane reacts with oxygen is:

  ${\text{CH}}_{\text{4}}{\text{+2O}}_{\text{2}}{\text{=2H}}_{\text{2}}{\text{O+CO}}_{\text{2}}$

Calculation:

As per given data $1.5\text{ mole}$ methane reacts with $3\text{ moles}$ of oxygen to give $3\text{ moles}$ water and 1 mole carbon dioxide.

1 mole water weighs $18\text{ g}$ .

So, 3 mole weighs $54\text{g}$ .

Conclusion:

The amount of water formed is $54\text{g}$ .