IDNLearn.com: Your reliable source for finding expert answers. Get the information you need from our community of experts who provide accurate and thorough answers to all your questions.
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].
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! IDNLearn.com has the answers you need. Thank you for visiting, and we look forward to helping you again soon.
