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!