Question

Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen gas is needed to react completely with 2.09 L of oxygen gas to produce water vapor? Answer in units of L.

Answer 1

Answer:

4.18L of hydrogen gas.

Explanation:

2H2(g) + O2(g) --------> 2H2O(g)

From the balanced reaction equation, 22.4 L of oxygen is required to react with 44.8L of hydrogen to produce water.

Therefore, 2.09L of oxygen will require 2.09 × 44.8/22.4 = 4.18L of hydrogen gas.

Recall that the volume ratio of 2:1 for hydrogen: oxygen is always maintained as typified above.

Answer 2

Answer:

$4.18LH_2$

Explanation:

The important thing to keep in mind when dealing with gases that are under the same conditions for pressure and temperature is that their mole ratio is equivalent to their volume ratio.

You can prove this by using the ideal gas law equation

P ⋅ V 1 = n 1 ⋅ R T⇒  the first gas at pressure  P  and temperature  T

P ⋅ V 2 = n 2 ⋅ R T ⇒  the second gas at pressure  P  and temperature  T

Divide these two equations to get

$\frac{PV_1}{PV_2} =\frac{n_1RT}{n_2RT}$

Therefore, you can say that

$\frac{n_1}{n_2} = \frac{V_1}{V_2}$ the mole ratio is equal to the volume ratio

The balanced chemical equation for your reaction looks like this

2 H 2(g] + O 2(g] → 2 H 2 O (g]

4.55 L O 2 ⋅ 2 L H 2 1 L O 2 = 9.10 L H 2

$= 2.09LO_2 .\frac{2LH_2}{1LO_2} \\=4.18LH_2$