If 22.5 L of Oxygen reacted with excess hydrogen, how many liters of water vapor could be produced?
If 22.5 L of oxygen react with excess hydrogen, how many liters of water vapor (gaseous water) could be produced
If the Temp and Pressure are the same use the mol ratio.
22.5 *(2mol/1mol)
22.5*2= 45L
To determine the number of liters of water vapor produced, we need to use the balanced equation for the reaction between oxygen and hydrogen. The balanced equation is:
2H2 + O2 → 2H2O
Based on the stoichiometry of the reaction, we can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
Now, we need to convert the given volume of oxygen (22.5 L) into moles using the ideal gas law equation:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since we are given the volume of oxygen and assuming the reaction is happening at standard temperature and pressure (STP), we can use the molar volume of gas at STP, which is 22.4 L/mol.
So, moles of oxygen = volume of oxygen / molar volume of oxygen at STP
= 22.5 L / 22.4 L/mol
≈ 1.004 moles of oxygen
According to the balanced equation, each mole of oxygen produces 2 moles of water. Therefore, multiplying the moles of oxygen by the molar ratio, we get:
moles of water = moles of oxygen × (2 moles of water / 1 mole of oxygen)
= 1.004 moles of oxygen × (2 moles of water / 1 mole of oxygen)
≈ 2.008 moles of water
Finally, we can convert the moles of water vapor to liters using the molar volume of gas at STP:
volume of water vapor = moles of water vapor × molar volume of water vapor at STP
= 2.008 moles of water × 22.4 L/mol
≈ 44.9656 L
Therefore, approximately 44.97 liters of water vapor could be produced.