How many moles of water vapour are formed when 10L of butane gas C4H10is burned in oxygen at STP
2C4H10 + 13O2 ==> 8CO2 + 10H2O
10 L C4H10 @ STP will produce 10L x 10/2 = 50 L H2O @ STP.
1 mol H2O will occupy 22.4 L. So how many mols are in 50 L?
Post your work if you get stuck.
2c4h10+13o2
Thank you I am very happy to get this answer
1mole of any gas equals the volume 22.4l. so 1mole=22.4l x =50l use chirs cross method and x=2.2mol
To find the number of moles of water vapor formed when 10L of butane gas (C4H10) is burned in oxygen at STP, we need to consider the balanced chemical equation for the combustion reaction. The balanced equation for the combustion of butane is:
2C4H10 + 13O2 → 8CO2 + 10H2O
From the balanced equation, we can see that for every 2 moles of butane (C4H10) burned, 10 moles of water (H2O) are formed.
We know that the volume of a gas is directly proportional to the number of moles of gas at constant temperature and pressure, according to the ideal gas law.
At STP (Standard Temperature and Pressure), the temperature is 273.15 K (0 degrees Celsius) and the pressure is 1 atmosphere (atm).
Since 1 mole of any gas occupies a volume of 22.4 liters at STP, we can calculate the number of moles of butane gas (C4H10) using the given volume and the ideal gas law:
Number of moles of butane (C4H10) = Volume of butane (C4H10) / Volume of 1 mole of any gas at STP
Number of moles of butane (C4H10) = 10 L / 22.4 L/mol
Now, we can determine the number of moles of water vapor formed by using the stoichiometry of the balanced equation:
Number of moles of water vapor = Number of moles of butane (C4H10) × (10 moles of water vapor / 2 moles of butane)
Number of moles of water vapor = (10 L / 22.4 L/mol) × (10 mol H2O / 2 mol C4H10)
Now, we can calculate the number of moles of water vapor formed.