Assuming all volume measurements are made at the same temperature and pressure, how many liters of water vapor can be produced when 8.23 liters of oxygen gas react with excess hydrogen gas? show all of the steps taken to solve this problem. balanced equation: 2h2 (g) + o2 (g) yields 2h2 o(g)

16.5 liter

Explanation:

Since the volume measurements are made at the same temperature and pressure, using Avogadro's prinicple, you conclude that the volume ratios are equal to the mole ratios.

That permit you to work directly with the same coefficients of the balanced equation.

This is how:

1) Balanced equation (given):

2) Mole and volume ratios:

3) Proportion:

1 liter O₂ / 2 liter H₂O = 8.23 liter O₂ / x

⇒ x = 8.23 liter O₂ × 2 liter H₂O / 1 liter O₂ = 16.46 liter H₂O

Rounding to three significant figures: 16.5 liter water vapor.

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