Answer :
Certainly! Let's walk through the steps to solve this problem:
### Step 1: Write the combustion reaction
The combustion reaction of acetylene (C₂H₂) with oxygen (O₂) is:
[tex]\[ 2 \text{C}_2\text{H}_2 + 5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \][/tex]
### Step 2: Calculate the molar mass of acetylene (C₂H₂)
The molar mass of acetylene (C₂H₂) is calculated as follows:
[tex]\[ \text{Molar mass of C}_2\text{H}_2 = (2 \times 12.01 \, \text{g/mol}) + (2 \times 1.008 \, \text{g/mol}) = 24.02 \, \text{g/mol} + 2.016 \, \text{g/mol} = 26.036 \, \text{g/mol} \][/tex]
### Step 3: Calculate the number of moles of acetylene (C₂H₂)
Given the mass of acetylene is 32.4 grams, we use its molar mass to find the number of moles:
[tex]\[ \text{Moles of C}_2\text{H}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{32.4 \, \text{g}}{26.036 \, \text{g/mol}} \approx 1.244 \, \text{mol} \][/tex]
### Step 4: Relate moles of acetylene to moles of carbon dioxide
The balanced equation shows that 2 moles of acetylene produce 4 moles of carbon dioxide:
[tex]\[ \text{2 moles of C}_2\text{H}_2 \rightarrow \text{4 moles of CO}_2 \][/tex]
Since 1 mole of C₂H₂ produces 2 moles of CO₂, we calculate:
[tex]\[ \text{Moles of CO}_2 = \text{Moles of C}_2\text{H}_2 \times 2 = 1.244 \, \text{mol} \times 2 = 2.488 \, \text{mol} \][/tex]
### Step 5: Convert the pressure from torr to atm
The given pressure is 607.9 torr. We convert it to atmospheres (atm) using the conversion factor [tex]\( 1 \, \text{atm} = 760 \, \text{torr} \)[/tex]:
[tex]\[ \text{Pressure in atm} = \frac{607.9 \, \text{torr}}{760 \, \text{torr/atm}} \approx 0.800 \, \text{atm} \][/tex]
### Step 6: Apply the Ideal Gas Law to find the volume of carbon dioxide (CO₂)
The Ideal Gas Law is [tex]\( PV = nRT \)[/tex]. Solving for [tex]\( V \)[/tex] (volume):
[tex]\[ V = \frac{nRT}{P} \][/tex]
Where:
- [tex]\( n \)[/tex] = moles of CO₂ = 2.488 mol
- [tex]\( R \)[/tex] = Ideal Gas Constant = 62.3637 \, \text{torr·L/(mol·K)}
- [tex]\( T \)[/tex] = Temperature in Kelvin = 276.9 K
- [tex]\( P \)[/tex] = Pressure in atm = 0.800 atm
Substitute the values into the equation:
[tex]\[ V = \frac{2.488 \, \text{mol} \times 62.3637 \, \text{torr·L/(mol·K)} \times 276.9 \, \text{K}}{0.800 \, \text{atm}} \][/tex]
Calculating this, we find:
[tex]\[ V \approx 53,732.50 \, \text{L} \][/tex]
Therefore, the volume of carbon dioxide (CO₂) produced is approximately 53,732.50 liters.
### Step 1: Write the combustion reaction
The combustion reaction of acetylene (C₂H₂) with oxygen (O₂) is:
[tex]\[ 2 \text{C}_2\text{H}_2 + 5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \][/tex]
### Step 2: Calculate the molar mass of acetylene (C₂H₂)
The molar mass of acetylene (C₂H₂) is calculated as follows:
[tex]\[ \text{Molar mass of C}_2\text{H}_2 = (2 \times 12.01 \, \text{g/mol}) + (2 \times 1.008 \, \text{g/mol}) = 24.02 \, \text{g/mol} + 2.016 \, \text{g/mol} = 26.036 \, \text{g/mol} \][/tex]
### Step 3: Calculate the number of moles of acetylene (C₂H₂)
Given the mass of acetylene is 32.4 grams, we use its molar mass to find the number of moles:
[tex]\[ \text{Moles of C}_2\text{H}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{32.4 \, \text{g}}{26.036 \, \text{g/mol}} \approx 1.244 \, \text{mol} \][/tex]
### Step 4: Relate moles of acetylene to moles of carbon dioxide
The balanced equation shows that 2 moles of acetylene produce 4 moles of carbon dioxide:
[tex]\[ \text{2 moles of C}_2\text{H}_2 \rightarrow \text{4 moles of CO}_2 \][/tex]
Since 1 mole of C₂H₂ produces 2 moles of CO₂, we calculate:
[tex]\[ \text{Moles of CO}_2 = \text{Moles of C}_2\text{H}_2 \times 2 = 1.244 \, \text{mol} \times 2 = 2.488 \, \text{mol} \][/tex]
### Step 5: Convert the pressure from torr to atm
The given pressure is 607.9 torr. We convert it to atmospheres (atm) using the conversion factor [tex]\( 1 \, \text{atm} = 760 \, \text{torr} \)[/tex]:
[tex]\[ \text{Pressure in atm} = \frac{607.9 \, \text{torr}}{760 \, \text{torr/atm}} \approx 0.800 \, \text{atm} \][/tex]
### Step 6: Apply the Ideal Gas Law to find the volume of carbon dioxide (CO₂)
The Ideal Gas Law is [tex]\( PV = nRT \)[/tex]. Solving for [tex]\( V \)[/tex] (volume):
[tex]\[ V = \frac{nRT}{P} \][/tex]
Where:
- [tex]\( n \)[/tex] = moles of CO₂ = 2.488 mol
- [tex]\( R \)[/tex] = Ideal Gas Constant = 62.3637 \, \text{torr·L/(mol·K)}
- [tex]\( T \)[/tex] = Temperature in Kelvin = 276.9 K
- [tex]\( P \)[/tex] = Pressure in atm = 0.800 atm
Substitute the values into the equation:
[tex]\[ V = \frac{2.488 \, \text{mol} \times 62.3637 \, \text{torr·L/(mol·K)} \times 276.9 \, \text{K}}{0.800 \, \text{atm}} \][/tex]
Calculating this, we find:
[tex]\[ V \approx 53,732.50 \, \text{L} \][/tex]
Therefore, the volume of carbon dioxide (CO₂) produced is approximately 53,732.50 liters.
