To determine how many liters of hydrogen and oxygen gas reacted to produce 7.84 liters of water vapor, we need to refer to the stoichiometric relationships given by the balanced chemical equation:
[tex]\[ 2 H_2 (g) + 1 O_2 (g) \rightarrow 2 H_2O (g) \][/tex]
From the balanced equation, we observe the following stoichiometric relationships:
- 2 moles of [tex]\( H_2 \)[/tex] produce 2 moles of [tex]\( H_2O \)[/tex].
- 1 mole of [tex]\( O_2 \)[/tex] is required to produce 2 moles of [tex]\( H_2O \)[/tex].
These relationships imply that the volumes of gases involved in the reaction at constant temperature and pressure will be directly proportional to the number of moles involved. This means:
- The volume of hydrogen gas needed will be equal to the volume of water vapor produced.
- The volume of oxygen gas needed will be half the volume of water vapor produced.
Given:
- 7.84 liters of [tex]\( H_2O \)[/tex] are produced.
We can determine the volumes of [tex]\( H_2 \)[/tex] and [tex]\( O_2 \)[/tex] as follows:
1. Volume of hydrogen gas ([tex]\( H_2 \)[/tex]) needed:
Since 2 moles of [tex]\( H_2 \)[/tex] react to produce 2 moles of [tex]\( H_2O \)[/tex], volume of [tex]\( H_2 \)[/tex] is equal to the volume of [tex]\( H_2O \)[/tex].
[tex]\[
\text{Volume of } H_2 = 7.84 \text{ liters}
\][/tex]
2. Volume of oxygen gas ([tex]\( O_2 \)[/tex]) needed:
Since 1 mole of [tex]\( O_2 \)[/tex] reacts to produce 2 moles of [tex]\( H_2O \)[/tex], the volume of [tex]\( O_2 \)[/tex] required is half the volume of [tex]\( H_2O \)[/tex].
[tex]\[
\text{Volume of } O_2 = \frac{7.84 \text{ liters}}{2} = 3.92 \text{ liters}
\][/tex]
Therefore, the reaction requires:
- 7.84 liters of [tex]\( H_2 \)[/tex] gas, and
- 3.92 liters of [tex]\( O_2 \)[/tex] gas to produce 7.84 liters of [tex]\( H_2O \)[/tex] vapor.
