Final answer:
To find the mass of water formed from 25.0L of propane, we use the balanced equation for propane combustion. By calculating the moles of propane and then the moles of water, the result is 80.44 grams of water produced.
Step-by-step explanation:
To calculate the mass of water formed from combusting 25.0L of propane (C₃H₈), we need to use the balanced chemical equation for the combustion of propane:
C3H8 + 5O2 → 3CO2 + 4H2O
Assuming the reaction goes to completion and occurs at standard temperature and pressure (STP), we can use the ideal gas law to determine the number of moles of propane. At STP, one mole of any gas occupies 22.4L. Therefore, we can find out the moles of propane:
Moles of propane = 25.0L / 22.4L/mole
= 1.116 moles propane
From the balanced equation, we see that one mole of propane yields four moles of water. Therefore:
Moles of water = 1.116 moles propane × 4 moles water/mole propane = 4.464 moles water
The molar mass of water (H2O) is approximately 18.015 g/mole. Therefore, the mass of water produced is:
Mass of water = 4.464 moles × 18.015 g/mole
= 80.44 g
The mass of water formed from 25.0L of propane is 80.44 grams.