At STP, if 4.20L of O 2 reacts with N2H4 how many liters of water vapor will be produced?
:You have to assume a reaction, and balance it.
N2H4 + O2>>N2 + 2H2O
looks like twice as much as 4.2L
To determine the volume of water vapor produced when 4.20 liters of O2 reacts with N2H4, we need to understand the balanced chemical equation for the reaction and use stoichiometry.
The balanced chemical equation for the reaction between O2 and N2H4 is:
2 N2H4 + O2 → 4 H2O + N2
From the equation, we can see that for every 1 mole of O2, 4 moles of water vapor (H2O) are produced. However, we need to convert the given volume of O2 to moles before applying stoichiometry.
To convert the volume of O2 to moles, we can use the ideal gas law:
PV = nRT
Where:
P = pressure (STP has a pressure of 1 atm)
V = volume (4.20 L)
n = number of moles (unknown)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (STP has a temperature of 273.15 K)
Rearranging the equation to solve for n, the number of moles:
n = PV / RT
Substituting the given values:
n = (1 atm) (4.20 L) / (0.0821 L·atm/mol·K) (273.15 K)
n ≈ 0.1701 mol
Now that we have the number of moles of O2, we can use stoichiometry to find the number of moles of water vapor produced.
From the balanced equation, we know that 1 mole of O2 reacts with 4 moles of water vapor. Therefore, if 0.1701 moles of O2 react:
Moles of water vapor = (0.1701 mol O2) × (4 mol H2O / 1 mol O2)
Moles of water vapor ≈ 0.6804 mol
Finally, we can convert moles of water vapor to volume using the ideal gas law, assuming STP conditions:
PV = nRT
Rearranging the equation:
V = nRT / P
Substituting the values:
V = (0.6804 mol) (0.0821 L·atm/mol·K) (273.15 K) / (1 atm)
V ≈ 12.33 L
Therefore, when 4.20 liters of O2 reacts with N2H4, approximately 12.33 liters of water vapor will be produced at standard temperature and pressure (STP).