Question
What is the maximum volume of butene mixture that can be obtained from 81 mg of 2-butanol at standard temperature and pressure (STP)?
Answers
GPT-4o mini
To determine the maximum volume of butene that can be obtained from 2-butanol, we need to consider the dehydrating reaction of 2-butanol to yield butene. The dehydration of 2-butanol can be represented by the following balanced chemical equation:
\[
\text{C}_4\text{H}_{10}\text{O} \rightarrow \text{C}_4\text{H}_8 + \text{H}_2\text{O}
\]
Here, one mole of 2-butanol (C\(_4\)H\(_{10}\)O) produces one mole of butene (C\(_4\)H\(_8\)) and one mole of water (H\(_2\)O).
### Step 1: Calculate the molar mass of 2-butanol
The molar mass of 2-butanol (C\(_4\)H\(_{10}\)O) can be calculated as follows:
- Carbon (C): \(12.01 \, \text{g/mol} \times 4 = 48.04 \, \text{g/mol}\)
- Hydrogen (H): \(1.008 \, \text{g/mol} \times 10 = 10.08 \, \text{g/mol}\)
- Oxygen (O): \(16.00 \, \text{g/mol} \times 1 = 16.00 \, \text{g/mol}\)
Total molar mass = \(48.04 + 10.08 + 16.00 = 74.12 \, \text{g/mol}\)
### Step 2: Calculate moles of 2-butanol
We have 81 mg of 2-butanol, which is equal to \(0.081 \, \text{g}\):
\[
\text{Moles of 2-butanol} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.081 \, \text{g}}{74.12 \, \text{g/mol}} \approx 0.001093 \, \text{mol}
\]
### Step 3: Calculate moles of butene produced
According to the balanced equation, 1 mole of 2-butanol produces 1 mole of butene. Therefore, the moles of butene produced will also be approximately \(0.001093 \, \text{mol}\).
### Step 4: Calculate volume of butene at STP
At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume of butene produced can be calculated as follows:
\[
\text{Volume of butene} = \text{moles of butene} \times 22.4 \, \text{L/mol} = 0.001093 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 0.0245 \, \text{L}
\]
### Step 5: Convert volume to milliliters
To convert liters to milliliters, multiply by 1000:
\[
\text{Volume of butene} \approx 0.0245 \, \text{L} \times 1000 \, \text{mL/L} = 24.5 \, \text{mL}
\]
Thus, the maximum volume of butene mixture that can be obtained from 81 mg of 2-butanol at standard temperature and pressure is approximately **24.5 mL**.
\[
\text{C}_4\text{H}_{10}\text{O} \rightarrow \text{C}_4\text{H}_8 + \text{H}_2\text{O}
\]
Here, one mole of 2-butanol (C\(_4\)H\(_{10}\)O) produces one mole of butene (C\(_4\)H\(_8\)) and one mole of water (H\(_2\)O).
### Step 1: Calculate the molar mass of 2-butanol
The molar mass of 2-butanol (C\(_4\)H\(_{10}\)O) can be calculated as follows:
- Carbon (C): \(12.01 \, \text{g/mol} \times 4 = 48.04 \, \text{g/mol}\)
- Hydrogen (H): \(1.008 \, \text{g/mol} \times 10 = 10.08 \, \text{g/mol}\)
- Oxygen (O): \(16.00 \, \text{g/mol} \times 1 = 16.00 \, \text{g/mol}\)
Total molar mass = \(48.04 + 10.08 + 16.00 = 74.12 \, \text{g/mol}\)
### Step 2: Calculate moles of 2-butanol
We have 81 mg of 2-butanol, which is equal to \(0.081 \, \text{g}\):
\[
\text{Moles of 2-butanol} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.081 \, \text{g}}{74.12 \, \text{g/mol}} \approx 0.001093 \, \text{mol}
\]
### Step 3: Calculate moles of butene produced
According to the balanced equation, 1 mole of 2-butanol produces 1 mole of butene. Therefore, the moles of butene produced will also be approximately \(0.001093 \, \text{mol}\).
### Step 4: Calculate volume of butene at STP
At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume of butene produced can be calculated as follows:
\[
\text{Volume of butene} = \text{moles of butene} \times 22.4 \, \text{L/mol} = 0.001093 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 0.0245 \, \text{L}
\]
### Step 5: Convert volume to milliliters
To convert liters to milliliters, multiply by 1000:
\[
\text{Volume of butene} \approx 0.0245 \, \text{L} \times 1000 \, \text{mL/L} = 24.5 \, \text{mL}
\]
Thus, the maximum volume of butene mixture that can be obtained from 81 mg of 2-butanol at standard temperature and pressure is approximately **24.5 mL**.