Ammonia is used as a fertilizer throughout the world. It is often produced by reacting nitrogen gas with hydrogen gas in a synthesis reaction. How much ammonia in grams can be produced from 9207 grams of nitrogen and 44 000 L of hydrogen?

N2 + 3H2 ==> 2NH3

This is a limiting reagent problem because amounts are given for BOTH reactants.
mols N2 = grams/molar mass
You don't give a pressure for H2 but at STP, 1 mol occupies 22.4L; therefore you have 44,000/22.4L = ? mols.

Convert mols N2 to mols NH3.
Convert mols H2 to mols NH3.
It is likely that these two values will not agree which means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value.

Now convert the smaller value to grams. g = mols x molar mass.

To determine the amount of ammonia that can be produced, we need to calculate the limiting reactant first. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.

Let's find the limiting reactant:

1. Determine the moles of nitrogen (N₂):
- Given: mass of nitrogen = 9207 grams
- Nitrogen molar mass: 28 grams/mole
- Moles of nitrogen = mass of nitrogen / molar mass of nitrogen
= 9207 g / 28 g/mol
≈ 329.53 moles

2. Determine the moles of hydrogen (H₂):
- Given: volume of hydrogen = 44000 L
- Hydrogen molar volume at STP (Standard Temperature and Pressure): 22.4 L/mol
- Moles of hydrogen = volume of hydrogen / molar volume of hydrogen
= 44000 L / 22.4 L/mol
≈ 1964.29 moles

3. Determine the mole ratio of nitrogen to ammonia in the balanced chemical equation.
The balanced chemical equation for the reaction: N₂ + 3H₂ -> 2NH₃
For every 1 mole of nitrogen, 2 moles of ammonia are produced.

4. Calculate the moles of ammonia that can be produced from the given amount of nitrogen:
- Moles of ammonia = (moles of nitrogen) x (moles of ammonia per mole of nitrogen)
= 329.53 moles x (2 moles NH₃ / 1 mole N₂)
≈ 659.06 moles

5. Finally, calculate the mass of ammonia:
- Given: Molar mass of ammonia (NH₃) = 17 grams/mole
- Mass of ammonia = moles of ammonia × molar mass of ammonia
= 659.06 moles × 17 g/mol
≈ 11193.02 grams

Therefore, approximately 11193.02 grams of ammonia can be produced from 9207 grams of nitrogen and 44000 L of hydrogen.