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Sagot :
Let's solve the problem step by step.
### Given:
1. Mass of ethanol: 209 grams
2. Volume of oxygen used at STP: 305 dm³
3. Volume of gas released at STP: 509 dm³
### Chemical Equation:
[tex]\[ 2 \text{CH}_3 \text{CH}_2 \text{OH}( \ell ) + 6 \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) + 6 \text{H}_2\text{O} (g) \][/tex]
### Definitions:
- Molar masses:
- Ethanol (CH₃CH₂OH): [tex]\( 46.07 \, \text{g/mol} \)[/tex]
- Oxygen (O₂): [tex]\( 32.00 \, \text{g/mol} \)[/tex]
- Carbon dioxide (CO₂): [tex]\( 44.01 \, \text{g/mol} \)[/tex]
- Water (H₂O): [tex]\( 18.02 \, \text{g/mol} \)[/tex]
- Volume of one mole of gas at STP: 22.414 dm³
### Step-by-Step Solutions:
#### a) Volume of Oxygen Used for Combustion at STP
1. Calculate the moles of ethanol ([tex]\(\text{CH}_3 \text{CH}_2 \text{OH}\)[/tex]):
[tex]\[ \text{Moles of ethanol} = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}} = \frac{209 \, \text{grams}}{46.07 \, \text{g/mol}} \approx 4.5366 \, \text{moles} \][/tex]
2. Using the stoichiometry of the reaction:
[tex]\[ 2 \text{moles of ethanol} \rightarrow 6 \text{moles of oxygen (O}_2\text{)} \][/tex]
[tex]\[ \text{Moles of oxygen used} = \left( \frac{6}{2} \right) \times \text{Moles of ethanol} \][/tex]
[tex]\[ \text{Moles of oxygen used} = 3 \times 4.5366 \approx 13.6097 \, \text{moles} \][/tex]
3. Calculate the volume of oxygen used at STP:
[tex]\[ \text{Volume of oxygen used at STP} = \text{Moles of oxygen used} \times \text{Volume of one mole of gas at STP} \][/tex]
[tex]\[ \text{Volume of oxygen used at STP} = 13.6097 \, \text{moles} \times 22.414 \, \text{dm}^3/\text{mole} \approx 305 \, \text{dm}^3 \][/tex]
So, the volume of oxygen used for combustion at STP is approximately [tex]\(305 \, \text{dm}^3\)[/tex].
#### b) Volume of Gas Released During Combustion at STP
1. Using the stoichiometry of the reaction:
[tex]\[ 2 \text{moles of ethanol} \rightarrow 4 \text{moles of CO}_2\text{(g)} + 6 \text{moles of H}_2 \text{O(g)} \][/tex]
Total moles of gas released:
[tex]\[ 4 \text{moles of CO}_2 + 6 \text{moles of H}_2 \text{O} = 10 \text{moles of gas} \][/tex]
2. Calculate the moles of gas released:
[tex]\[ \text{Moles of gas released} = \left( \frac{10}{2} \right) \times \text{Moles of ethanol} \][/tex]
[tex]\[ \text{Moles of gas released} = 5 \times 4.5366 \approx 22.6829 \, \text{moles} \][/tex]
3. Calculate the volume of gas released at STP:
[tex]\[ \text{Volume of gas released at STP} = \text{Moles of gas released} \times \text{Volume of one mole of gas at STP} \][/tex]
[tex]\[ \text{Volume of gas released at STP} = 22.6829 \, \text{moles} \times 22.414 \, \text{dm}^3/\text{mole} \approx 509 \, \text{dm}^3 \][/tex]
So, the volume of gas released during combustion at STP is approximately [tex]\(509 \, \text{dm}^3\)[/tex].
### Summary:
a) The volume of oxygen used for combustion at STP is approximately [tex]\(305 \, \text{dm}^3\)[/tex].
b) The volume of gas released during combustion at STP is approximately [tex]\(509 \, \text{dm}^3\)[/tex].
### Given:
1. Mass of ethanol: 209 grams
2. Volume of oxygen used at STP: 305 dm³
3. Volume of gas released at STP: 509 dm³
### Chemical Equation:
[tex]\[ 2 \text{CH}_3 \text{CH}_2 \text{OH}( \ell ) + 6 \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) + 6 \text{H}_2\text{O} (g) \][/tex]
### Definitions:
- Molar masses:
- Ethanol (CH₃CH₂OH): [tex]\( 46.07 \, \text{g/mol} \)[/tex]
- Oxygen (O₂): [tex]\( 32.00 \, \text{g/mol} \)[/tex]
- Carbon dioxide (CO₂): [tex]\( 44.01 \, \text{g/mol} \)[/tex]
- Water (H₂O): [tex]\( 18.02 \, \text{g/mol} \)[/tex]
- Volume of one mole of gas at STP: 22.414 dm³
### Step-by-Step Solutions:
#### a) Volume of Oxygen Used for Combustion at STP
1. Calculate the moles of ethanol ([tex]\(\text{CH}_3 \text{CH}_2 \text{OH}\)[/tex]):
[tex]\[ \text{Moles of ethanol} = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}} = \frac{209 \, \text{grams}}{46.07 \, \text{g/mol}} \approx 4.5366 \, \text{moles} \][/tex]
2. Using the stoichiometry of the reaction:
[tex]\[ 2 \text{moles of ethanol} \rightarrow 6 \text{moles of oxygen (O}_2\text{)} \][/tex]
[tex]\[ \text{Moles of oxygen used} = \left( \frac{6}{2} \right) \times \text{Moles of ethanol} \][/tex]
[tex]\[ \text{Moles of oxygen used} = 3 \times 4.5366 \approx 13.6097 \, \text{moles} \][/tex]
3. Calculate the volume of oxygen used at STP:
[tex]\[ \text{Volume of oxygen used at STP} = \text{Moles of oxygen used} \times \text{Volume of one mole of gas at STP} \][/tex]
[tex]\[ \text{Volume of oxygen used at STP} = 13.6097 \, \text{moles} \times 22.414 \, \text{dm}^3/\text{mole} \approx 305 \, \text{dm}^3 \][/tex]
So, the volume of oxygen used for combustion at STP is approximately [tex]\(305 \, \text{dm}^3\)[/tex].
#### b) Volume of Gas Released During Combustion at STP
1. Using the stoichiometry of the reaction:
[tex]\[ 2 \text{moles of ethanol} \rightarrow 4 \text{moles of CO}_2\text{(g)} + 6 \text{moles of H}_2 \text{O(g)} \][/tex]
Total moles of gas released:
[tex]\[ 4 \text{moles of CO}_2 + 6 \text{moles of H}_2 \text{O} = 10 \text{moles of gas} \][/tex]
2. Calculate the moles of gas released:
[tex]\[ \text{Moles of gas released} = \left( \frac{10}{2} \right) \times \text{Moles of ethanol} \][/tex]
[tex]\[ \text{Moles of gas released} = 5 \times 4.5366 \approx 22.6829 \, \text{moles} \][/tex]
3. Calculate the volume of gas released at STP:
[tex]\[ \text{Volume of gas released at STP} = \text{Moles of gas released} \times \text{Volume of one mole of gas at STP} \][/tex]
[tex]\[ \text{Volume of gas released at STP} = 22.6829 \, \text{moles} \times 22.414 \, \text{dm}^3/\text{mole} \approx 509 \, \text{dm}^3 \][/tex]
So, the volume of gas released during combustion at STP is approximately [tex]\(509 \, \text{dm}^3\)[/tex].
### Summary:
a) The volume of oxygen used for combustion at STP is approximately [tex]\(305 \, \text{dm}^3\)[/tex].
b) The volume of gas released during combustion at STP is approximately [tex]\(509 \, \text{dm}^3\)[/tex].
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