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What volume of chlorine (Cl[tex]\(_2\)[/tex]) is necessary to prepare 38.6 liters of hydrogen chloride (HCl) if an excess of hydrogen (H[tex]\(_2\)[/tex]) is used? All gases are measured at 100°C and 1 atmosphere pressure.

Answer :

To solve the problem of finding the volume of chlorine ([tex]\( \text{Cl}_2 \)[/tex]) needed to prepare 38.6 liters of hydrogen chloride ([tex]\( \text{HCl} \)[/tex]) when an excess of hydrogen ([tex]\( \text{H}_2 \)[/tex]) is used, we must refer to the chemical reaction between chlorine and hydrogen:

[tex]\[ \text{Cl}_2 + \text{H}_2 \rightarrow 2\text{HCl} \][/tex]

This balanced equation shows that one molecule of chlorine ([tex]\( \text{Cl}_2 \)[/tex]) reacts with one molecule of hydrogen ([tex]\( \text{H}_2 \)[/tex]) to produce two molecules of hydrogen chloride ([tex]\( \text{HCl} \)[/tex]).

Let's go through the solution step-by-step:

1. Identify the volume of HCl produced:
According to the problem, we want to produce 38.6 liters of [tex]\( \text{HCl} \)[/tex].

2. Use stoichiometry to find the volume of Cl2 needed:
The stoichiometry of the reaction tells us the volume ratio between the reactants and products. It shows that:

- 1 volume of [tex]\( \text{Cl}_2 \)[/tex] produces 2 volumes of [tex]\( \text{HCl} \)[/tex].

3. Calculate the volume of Cl2 required:
To produce 38.6 liters of [tex]\( \text{HCl} \)[/tex], we determine the necessary volume of [tex]\( \text{Cl}_2 \)[/tex] using the 1:2 ratio. Since twice the volume of [tex]\( \text{HCl} \)[/tex] is produced per volume of [tex]\( \text{Cl}_2 \)[/tex],
you need half the volume of [tex]\( \text{HCl} \)[/tex] in [tex]\( \text{Cl}_2 \)[/tex]:

[tex]\[
\text{Volume of } \text{Cl}_2 = \frac{\text{Volume of } \text{HCl}}{2} = \frac{38.6}{2} = 19.3 \text{ liters}
\][/tex]

Thus, 19.3 liters of chlorine ([tex]\( \text{Cl}_2 \)[/tex]) is required to produce 38.6 liters of hydrogen chloride ([tex]\( \text{HCl} \)[/tex]).

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