Given the reaction N2 + 3H2 -->2NH3, what volume of hydrogen is necessary to react with five liters of nitrogen to produce ammonia, assuming constant temperature and pressure?

Reactant addition will be determined by stoichiometry, assuming full reaction and no excess of reactants.

Since volume is proportional to the number of moles in the ideal gas relation (PV=nRT), the addition of reactant by volume will be based on its stoichiometric coefficient, assuming ideal gas behavior. This means that the reaction between 1 mole of reactant A and 3 mole of reactant B is the same as X L of reactant A with 3X L of reactant B.

According to the stoichiometry of the reaction, for every one mole of nitrogen (N2), three moles of hydrogen (H2) are required to produce two moles of ammonia (NH3). Since the reaction is occurring at constant temperature and pressure, we can use the ideal gas law to find the volume of hydrogen needed.

First, we need to convert the volume of nitrogen given in liters to moles using the ideal gas law equation:

[ n = \frac{PV}{RT} ]

Where:

• ( P ) is the pressure (assuming constant)
• ( V ) is the volume of nitrogen (given as 5 liters)
• ( R ) is the ideal gas constant
• ( T ) is the temperature (assuming constant)

Next, we use the stoichiometry of the reaction to determine the mole ratio of hydrogen to nitrogen:

[ \text{Mole ratio of } H2 : N2 = 3 : 1 ]

Finally, we use the mole ratio to find the volume of hydrogen needed:

[ \text{Volume of } H2 = \frac{3}{1} \times \text{moles of } N2 \times \frac{RT}{P} ]

Substitute the given values to calculate the volume of hydrogen.