C2H4 n 0,4, m 11,2 V 22,4, M

To analyze the given data for \( \text{C}_2\text{H}_4 \) (ethene orethylene), we can identify the parameters you provided: - \( n \): number of moles - \( V \): volume - \( m \): mass - \( M \): molar mass 1. **Given Data**: - \( n = 0.4 \, \text{mol} \) - \( V = 22.4 \, \text{L} \) (at STP, where 1 mole of a gas occupies approximately 22.4 L) - \( m = 11.2 \, \text{g} \) 2. **Molar Mass Calculation**: The molar mass \( M \) of ethene (\( \text{C}_2\text{H}_4 \)) can be calculated using the formula: \[ M = \frac{m}{n} \] where \( m \) is the mass in grams and \( n \) is the number of moles. Plugging in the values: \[ M = \frac{11.2 \, \text{g}}{0.4 \, \text{mol}} = 28 \, \text{g/mol} \] 3. **Confirm Molar Mass**: Ethene (\( \text{C}_2\text{H}_4 \)) has a molar mass calculated as follows: - Carbon (C) has a molar mass of approximately 12.01 g/mol. For 2 carbons: \[ 2 \times 12.01 = 24.02 \, \text{g/mol} \] - Hydrogen (H) has a molar mass of approximately 1.008 g/mol. For 4 hydrogens: \[ 4 \times 1.008 = 4.032 \, \text{g/mol} \] - Therefore, the total molar mass of \( \text{C}_2\text{H}_4 \) is: \[ 24.02 + 4.032 = 28.052 \, \text{g/mol} \approx 28 \, \text{g/mol} \] Thus, your calculated values are consistent. The molar mass \( M \) of ethene (\( \text{C}_2\text{H}_4 \)) is approximately 28 g/mol. Let me know if you need further assistance with any other concepts or calculations!