Answer :
Let's solve the given problem step by step.
The balanced chemical equation for the combustion of carbon disulfide ([tex]\(CS_2\)[/tex]) in oxygen ([tex]\(O_2\)[/tex]) is as follows:
[tex]\[ CS_2(g) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g) \][/tex]
We need to calculate the volumes of [tex]\(O_2\)[/tex], [tex]\(CO_2\)[/tex], and [tex]\(SO_2\)[/tex] involved in the reaction when [tex]\(3 \, \text{dm}^3\)[/tex] of [tex]\(CS_2\)[/tex] reacts completely.
### Part 6.1: Volume of [tex]\(O_2\)[/tex] used
From the balanced equation, we see that 1 volume of [tex]\(CS_2\)[/tex] reacts with 3 volumes of [tex]\(O_2\)[/tex]. Therefore, the volume of [tex]\(O_2\)[/tex] used will be three times the volume of [tex]\(CS_2\)[/tex].
Given:
- Volume of [tex]\(CS_2 = 3 \, \text{dm}^3\)[/tex]
Using the stoichiometric ratio from the balanced equation:
- Volume of [tex]\(O_2\)[/tex] used = Volume of [tex]\(CS_2\)[/tex] [tex]\(\times 3\)[/tex]
- Volume of [tex]\(O_2\)[/tex] used = [tex]\(3 \, \text{dm}^3 \times 3 = 9 \, \text{dm}^3\)[/tex]
### Part 6.2: Volume of [tex]\(CO_2\)[/tex] and [tex]\(SO_2\)[/tex] made
From the balanced equation, we also see that:
- 1 volume of [tex]\(CS_2\)[/tex] produces 1 volume of [tex]\(CO_2\)[/tex].
- 1 volume of [tex]\(CS_2\)[/tex] produces 2 volumes of [tex]\(SO_2\)[/tex].
Therefore,
- Volume of [tex]\(CO_2\)[/tex] produced = Volume of [tex]\(CS_2 \, (\text{dm}^3}) = 3 \, \text{dm}^3 - Volume of \(SO_2\)[/tex] produced = Volume of [tex]\(CS_2\)[/tex] [tex]\(\times 2\)[/tex] = [tex]\(3 \, \text{dm}^3 \times 2 = 6 \, \text{dm}^3\)[/tex]
### Summary
When [tex]\(3 \, \text{dm}^3\)[/tex] of [tex]\(CS_2\)[/tex] reacts completely:
- Volume of [tex]\(O_2\)[/tex] used is [tex]\(9 \, \text{dm}^3\)[/tex].
- Volume of [tex]\(CO_2\)[/tex] produced is [tex]\(3 \, \text{dm}^3\)[/tex].
- Volume of [tex]\(SO_2\)[/tex] produced is [tex]\(6 \, \text{dm}^3\)[/tex].
Thus, the answers are:
1. Volume of [tex]\(O_2\)[/tex] used: [tex]\(9 \, \text{dm}^3\)[/tex]
2. Volume of [tex]\(CO_2\)[/tex] produced: [tex]\(3 \, \text{dm}^3\)[/tex]
3. Volume of [tex]\(SO_2\)[/tex] produced: [tex]\(6 \, \text{dm}^3\)[/tex]
The balanced chemical equation for the combustion of carbon disulfide ([tex]\(CS_2\)[/tex]) in oxygen ([tex]\(O_2\)[/tex]) is as follows:
[tex]\[ CS_2(g) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g) \][/tex]
We need to calculate the volumes of [tex]\(O_2\)[/tex], [tex]\(CO_2\)[/tex], and [tex]\(SO_2\)[/tex] involved in the reaction when [tex]\(3 \, \text{dm}^3\)[/tex] of [tex]\(CS_2\)[/tex] reacts completely.
### Part 6.1: Volume of [tex]\(O_2\)[/tex] used
From the balanced equation, we see that 1 volume of [tex]\(CS_2\)[/tex] reacts with 3 volumes of [tex]\(O_2\)[/tex]. Therefore, the volume of [tex]\(O_2\)[/tex] used will be three times the volume of [tex]\(CS_2\)[/tex].
Given:
- Volume of [tex]\(CS_2 = 3 \, \text{dm}^3\)[/tex]
Using the stoichiometric ratio from the balanced equation:
- Volume of [tex]\(O_2\)[/tex] used = Volume of [tex]\(CS_2\)[/tex] [tex]\(\times 3\)[/tex]
- Volume of [tex]\(O_2\)[/tex] used = [tex]\(3 \, \text{dm}^3 \times 3 = 9 \, \text{dm}^3\)[/tex]
### Part 6.2: Volume of [tex]\(CO_2\)[/tex] and [tex]\(SO_2\)[/tex] made
From the balanced equation, we also see that:
- 1 volume of [tex]\(CS_2\)[/tex] produces 1 volume of [tex]\(CO_2\)[/tex].
- 1 volume of [tex]\(CS_2\)[/tex] produces 2 volumes of [tex]\(SO_2\)[/tex].
Therefore,
- Volume of [tex]\(CO_2\)[/tex] produced = Volume of [tex]\(CS_2 \, (\text{dm}^3}) = 3 \, \text{dm}^3 - Volume of \(SO_2\)[/tex] produced = Volume of [tex]\(CS_2\)[/tex] [tex]\(\times 2\)[/tex] = [tex]\(3 \, \text{dm}^3 \times 2 = 6 \, \text{dm}^3\)[/tex]
### Summary
When [tex]\(3 \, \text{dm}^3\)[/tex] of [tex]\(CS_2\)[/tex] reacts completely:
- Volume of [tex]\(O_2\)[/tex] used is [tex]\(9 \, \text{dm}^3\)[/tex].
- Volume of [tex]\(CO_2\)[/tex] produced is [tex]\(3 \, \text{dm}^3\)[/tex].
- Volume of [tex]\(SO_2\)[/tex] produced is [tex]\(6 \, \text{dm}^3\)[/tex].
Thus, the answers are:
1. Volume of [tex]\(O_2\)[/tex] used: [tex]\(9 \, \text{dm}^3\)[/tex]
2. Volume of [tex]\(CO_2\)[/tex] produced: [tex]\(3 \, \text{dm}^3\)[/tex]
3. Volume of [tex]\(SO_2\)[/tex] produced: [tex]\(6 \, \text{dm}^3\)[/tex]