Question
what volume is occupied by 55 g of methane, CH4 (g) at stp
Answers
GPT-4o mini
To calculate the volume occupied by 55 g of methane (CH₄) at standard temperature and pressure (STP), we can use the ideal gas law and the molar volume of a gas at STP.
1. **Determine the molar mass of methane (CH₄):**
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol × 4 = 4.04 g/mol
- Molar mass of CH₄ = 12.01 g/mol + 4.04 g/mol = 16.05 g/mol
2. **Calculate the number of moles of methane in 55 g:**
\[
\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{55 \, \text{g}}{16.05 \, \text{g/mol}} \approx 3.43 \, \text{mol}
\]
3. **At STP, one mole of gas occupies 22.4 liters.**
4. **Calculate the volume occupied by 3.43 moles of CH₄:**
\[
\text{Volume} = \text{number of moles} \times \text{molar volume} = 3.43 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 76.91 \, \text{L}
\]
Therefore, the volume occupied by 55 g of methane at STP is approximately **76.91 liters**.
1. **Determine the molar mass of methane (CH₄):**
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol × 4 = 4.04 g/mol
- Molar mass of CH₄ = 12.01 g/mol + 4.04 g/mol = 16.05 g/mol
2. **Calculate the number of moles of methane in 55 g:**
\[
\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{55 \, \text{g}}{16.05 \, \text{g/mol}} \approx 3.43 \, \text{mol}
\]
3. **At STP, one mole of gas occupies 22.4 liters.**
4. **Calculate the volume occupied by 3.43 moles of CH₄:**
\[
\text{Volume} = \text{number of moles} \times \text{molar volume} = 3.43 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 76.91 \, \text{L}
\]
Therefore, the volume occupied by 55 g of methane at STP is approximately **76.91 liters**.