Answer :
Answer:
The combustion reaction of ethylene (C2H4) is:
C2H4 + 3O2 → 2CO2 + 2H2O
From the reaction, 1 mole of ethylene produces 2 moles of water.
First, let's calculate the number of moles of ethylene:
50 mL × (1 mole/22.4 L) = 0.00223 moles (assuming ideal.
The combustion reaction of ethylene (C2H4) is:
C2H4 + 3O2 → 2CO2 + 2H2O
From the reaction, 1 mole of ethylene produces 2 moles of water.
First, let's calculate the number of moles of ethylene:
50 mL = 0.05 L ( converting milliliters to liters)
Using the molar volume of an ideal gas (22.4 L/mol), we get:
0.05 L / 22.4 L/mol = 0.00223 mol (approx)
Now, multiply the number of moles of ethylene by the mole ratio of water to ethylene (2:1):
0.00223 mol × 2 mol H2O / 1 mol C2H4 = 0.00446 mol H2O
Finally, calculate the volume of water produced:
0.00446 mol × 18.02 g/mol (molar mass of water) = 0.0803 g
Using the density of water (approximately 1 g/mL), we get:
0.0803 g / 1 g/mL = 0.0803 mL (or approximately 80.3 μL)
So, the combustion of 50 mL of ethylene produces approximately 80.3 μL of water.
