Answer :
To solve the given problem step-by-step, let's analyze the chemical reaction and perform the necessary calculations. The reaction is as follows:
[tex]\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g) \][/tex]
### Part a:
#### i. Finding the volume of hydrogen chloride gas produced
1. Identify the initial volumes:
- Volume of hydrogen ([tex]\(\text{H}_2\)[/tex]) = 40 cm³
- Volume of chlorine ([tex]\(\text{Cl}_2\)[/tex]) = 10 cm³
2. Use the stoichiometric coefficients from the balanced chemical equation:
- According to the equation, 1 volume of [tex]\(\text{H}_2\)[/tex] reacts with 1 volume of [tex]\(\text{Cl}_2\)[/tex] to produce 2 volumes of [tex]\(\text{HCl}\)[/tex].
3. Determine the limiting reactant:
- Compare the given volumes of [tex]\(\text{H}_2\)[/tex] and [tex]\(\text{Cl}_2\)[/tex].
- Since 40 cm³ of [tex]\(\text{H}_2\)[/tex] would require 40 cm³ of [tex]\(\text{Cl}_2\)[/tex] to react completely and we have only 10 cm³ of [tex]\(\text{Cl}_2\)[/tex], [tex]\(\text{Cl}_2\)[/tex] is the limiting reactant.
4. Calculate the volume of [tex]\(\text{HCl}\)[/tex] produced:
- Since [tex]\(\text{Cl}_2\)[/tex] is the limiting reactant with 10 cm³, it will determine the amount of [tex]\(\text{HCl}\)[/tex] produced.
- From the equation, 1 volume of [tex]\(\text{Cl}_2\)[/tex] produces 2 volumes of [tex]\(\text{HCl}\)[/tex].
- Therefore, 10 cm³ of [tex]\(\text{Cl}_2\)[/tex] will produce [tex]\(2 \times 10\)[/tex] cm³ = 20 cm³ of [tex]\(\text{HCl}\)[/tex].
Thus, the volume of hydrogen chloride gas produced is 20 cm³.
#### ii. Finding the volume of any excess reactant gas after reaction
1. Calculate the excess reactant:
- Initially, we have 40 cm³ of [tex]\(\text{H}_2\)[/tex].
- Since [tex]\(\text{Cl}_2\)[/tex] is the limiting reactant and only 10 cm³ of [tex]\(\text{Cl}_2\)[/tex] is available, it will react with 10 cm³ of [tex]\(\text{H}_2\)[/tex].
2. Determine the remaining volume of [tex]\(\text{H}_2\)[/tex]:
- [tex]\(40 \text{ cm}³ - 10 \text{ cm}³ = 30 \text{ cm}³\)[/tex]
Therefore, the volume of the excess hydrogen gas after the reaction is 30 cm³.
### Part b:
#### Law and assumptions:
The law assumed in these calculations is Gay-Lussac's law of gaseous volumes.
Gay-Lussac's law of gaseous volumes states that, during a chemical reaction between gases, the volumes of the reacting gases and the products (if gaseous) are in simple ratios to the coefficients in the balanced chemical equation, under the same conditions of temperature and pressure.
Assumptions being adopted:
1. The reaction goes to completion.
2. The temperature and pressure remain constant during the reaction.
3. The volumes of the gases are measured under the same conditions of temperature and pressure.
Therefore, using Gay-Lussac's law, we can determine the stoichiometric volumes of gases involved in the reaction.
[tex]\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g) \][/tex]
### Part a:
#### i. Finding the volume of hydrogen chloride gas produced
1. Identify the initial volumes:
- Volume of hydrogen ([tex]\(\text{H}_2\)[/tex]) = 40 cm³
- Volume of chlorine ([tex]\(\text{Cl}_2\)[/tex]) = 10 cm³
2. Use the stoichiometric coefficients from the balanced chemical equation:
- According to the equation, 1 volume of [tex]\(\text{H}_2\)[/tex] reacts with 1 volume of [tex]\(\text{Cl}_2\)[/tex] to produce 2 volumes of [tex]\(\text{HCl}\)[/tex].
3. Determine the limiting reactant:
- Compare the given volumes of [tex]\(\text{H}_2\)[/tex] and [tex]\(\text{Cl}_2\)[/tex].
- Since 40 cm³ of [tex]\(\text{H}_2\)[/tex] would require 40 cm³ of [tex]\(\text{Cl}_2\)[/tex] to react completely and we have only 10 cm³ of [tex]\(\text{Cl}_2\)[/tex], [tex]\(\text{Cl}_2\)[/tex] is the limiting reactant.
4. Calculate the volume of [tex]\(\text{HCl}\)[/tex] produced:
- Since [tex]\(\text{Cl}_2\)[/tex] is the limiting reactant with 10 cm³, it will determine the amount of [tex]\(\text{HCl}\)[/tex] produced.
- From the equation, 1 volume of [tex]\(\text{Cl}_2\)[/tex] produces 2 volumes of [tex]\(\text{HCl}\)[/tex].
- Therefore, 10 cm³ of [tex]\(\text{Cl}_2\)[/tex] will produce [tex]\(2 \times 10\)[/tex] cm³ = 20 cm³ of [tex]\(\text{HCl}\)[/tex].
Thus, the volume of hydrogen chloride gas produced is 20 cm³.
#### ii. Finding the volume of any excess reactant gas after reaction
1. Calculate the excess reactant:
- Initially, we have 40 cm³ of [tex]\(\text{H}_2\)[/tex].
- Since [tex]\(\text{Cl}_2\)[/tex] is the limiting reactant and only 10 cm³ of [tex]\(\text{Cl}_2\)[/tex] is available, it will react with 10 cm³ of [tex]\(\text{H}_2\)[/tex].
2. Determine the remaining volume of [tex]\(\text{H}_2\)[/tex]:
- [tex]\(40 \text{ cm}³ - 10 \text{ cm}³ = 30 \text{ cm}³\)[/tex]
Therefore, the volume of the excess hydrogen gas after the reaction is 30 cm³.
### Part b:
#### Law and assumptions:
The law assumed in these calculations is Gay-Lussac's law of gaseous volumes.
Gay-Lussac's law of gaseous volumes states that, during a chemical reaction between gases, the volumes of the reacting gases and the products (if gaseous) are in simple ratios to the coefficients in the balanced chemical equation, under the same conditions of temperature and pressure.
Assumptions being adopted:
1. The reaction goes to completion.
2. The temperature and pressure remain constant during the reaction.
3. The volumes of the gases are measured under the same conditions of temperature and pressure.
Therefore, using Gay-Lussac's law, we can determine the stoichiometric volumes of gases involved in the reaction.
