Consider the following equation:

4 NH3(g) + 5 O2(g)= 4 NO(g) + 6 H2O(s)

a) HOw many liters of oxygen are required to react with 205 L NH3? Both gases are STP.

b)How many grams of water vapor can be produced from 25 L NH3 if both gases are STP?

c) How many liters of NO can be produced when 25 L O2 are reacted with 25 L NH3? All gases are at the same temperature and pressure.

I am not understanding how to do this at all. I am stuck completely. Can you please help me.

When working with gases like this you can take a short cut and use L as if they were mols. Then just use the coefficients to change anything you have to anything you want.

a.
205L NH3 x (5 mols O2/4 mols NH3) = ? L NH3
You try the others.

Sure, I'd be happy to help you step by step with these questions.

a) To determine the number of liters of oxygen required, you need to use the stoichiometry of the balanced equation. The mole ratio between NH3 and O2 is 4:5. Therefore, for every 4 moles of NH3, you need 5 moles of O2.

To solve the problem, you can follow these steps:

1. Convert the given volume of NH3 to moles at STP (Standard Temperature and Pressure). Remember that at STP, one mole of any ideal gas occupies 22.4 L.
- 205 L NH3 / 22.4 L/mol = 9.15 mol NH3

2. Use the mole ratio from the balanced equation to determine the moles of O2 required.
- 9.15 mol NH3 × (5 mol O2 / 4 mol NH3) = 11.44 mol O2

3. Convert the moles of O2 to volume at STP.
- 11.44 mol O2 × 22.4 L/mol = 256.26 L O2

Therefore, you would require 256.26 liters of oxygen to react with 205 liters of NH3.

b) To determine the grams of water produced, you need to use the same stoichiometry principle. The mole ratio between NH3 and H2O is 4:6, meaning 4 moles of NH3 react to form 6 moles of H2O.

Here's how you can calculate the answer:

1. Convert the given volume of NH3 to moles at STP.
- 25 L NH3 / 22.4 L/mol = 1.12 mol NH3

2. Use the mole ratio to determine the moles of H2O produced.
- 1.12 mol NH3 × (6 mol H2O / 4 mol NH3) = 1.68 mol H2O

3. Convert the moles of H2O to grams using the molar mass of water (18.0153 g/mol).
- 1.68 mol H2O × 18.0153 g/mol = 30.27 g H2O

Therefore, you would produce 30.27 grams of water vapor from 25 liters of NH3.

c) To determine the liters of NO produced, you need to compare the reactant volumes of O2 and NH3. According to the balanced equation, 5 liters of O2 react with 4 liters of NH3 to produce 4 liters of NO.

Here's how you can solve the problem:

Since both gases are at the same temperature and pressure, we can say that the volumes of gases are directly proportional to the number of moles. Therefore, you can assume that 25 liters of O2 and 25 liters of NH3 have an equal number of moles.

Using this information, you can conclude that 25 liters of O2 will react with 25 liters of NH3 to produce 25 liters of NO.

I hope this helps clarify the process for you. Let me know if you have any further questions!