Question

5. In one method of manufacturing nitric acid, ammonia is oxidized to nitrogen monoxide and water:

How many grams of water will be produced in a reaction of 2800. L. of nitrogen trihydride?
4NH₂(g) + 50.(g) → 4NO(g) + 6H₂O)
03/16/2023 12:03

Answer 1

The balanced chemical equation for the reaction is:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

From the equation, we can see that for every 4 moles of NH3 reacted, 6 moles of H2O are produced. Therefore, we can use the following conversion factors:

4 moles NH3 → 6 moles H2O
1 mole NH3 → 6/4 moles H2O
1 mole NH3 → 1.5 moles H2O

To convert the volume of NH3 gas to moles, we need to use the ideal gas law:

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.

Assuming standard temperature and pressure (STP), which is 0°C and 1 atm, the volume of 2800 L of NH3 gas contains:

n = PV/RT = (1 atm)(2800 L)/(0.08206 L·atm/mol·K)(273 K) ≈ 1214.5 moles NH3

Using the conversion factor above, we can calculate the number of moles of H2O produced:

n(H2O) = n(NH3) × (1.5 moles H2O/1 mole NH3) ≈n(H2O) = 1214.5 moles NH3 × (1.5 moles H2O/1 mole NH3) ≈ 1821.75 moles H2O

Finally, we can convert the number of moles of H2O to grams using the molar mass of water:

m(H2O) = n(H2O) × M(H2O) = 1821.75 moles H2O × 18.015 g/mol ≈ 32,787 g or 32.787 kg

Therefore, approximately 32.787 kg or 32,787 g of water will be produced in the reaction of 2800 L of NH3 gas.