A mixture of nitrogen gas and hydrogen gas reacts in a closed container to form ammonia

gas. The reaction ceases before either reactant is completely consumed. At this stage the
container has 3.0 mol each of all three gases; nitrogen, hydrogen and ammonia. How many
moles of hydrogen and nitrogen were present initially?

i'm not even sure how to start this question honestly.

I don't think this problem can be worked without knowing Keq. You could calculate Keq if you had the volume.

N2 + 3H2 ==> 2NH3

Keq = (NH3)^3/(N2)(H2)^3.
Substitute and solve for Keq.

To solve this question, we will use the concept of stoichiometry and the ideal gas law. Here's a step-by-step approach to finding the number of moles of hydrogen and nitrogen initially present:

1. Write the balanced chemical equation for the reaction:
N2(g) + 3H2(g) -> 2NH3(g)

This equation tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

2. Calculate the moles of ammonia formed:
Given that the container has 3.0 moles of ammonia, we can assume that the reaction has gone to completion. Therefore, the number of moles of ammonia formed is equal to the number of moles of nitrogen initially present. So, there were 3.0 moles of nitrogen initially.

3. Apply the stoichiometry of the reaction to find the moles of hydrogen:
From the balanced equation, we know that the molar ratio of hydrogen to nitrogen is 3:1. Since 3 moles of hydrogen react with 1 mole of nitrogen, the number of moles of hydrogen initially present would be three times the number of moles of nitrogen. Thus, the number of moles of hydrogen initially present is 3.0 * 3 = 9.0 moles.

Therefore, the initial quantities of hydrogen and nitrogen were 9.0 moles and 3.0 moles, respectively.