Nitrogen gas reacts with hydrogen gas to produce ammonia as outlined in the following equation:      N2 (g) + 3 H2 (g)  ->   2 NH3 (g)If a reaction vessel contains 28 grams of N2 (g) and 28 grams of H2 (g) how many moles of ammonia gas are produced once the reaction is complete?

To solve this problem, we need to follow these steps:

1. First, we need to determine the number of moles of N2 and H2. We do this by dividing the mass of each gas by its molar mass. The molar mass of N2 is approximately 28 g/mol and the molar mass of H2 is approximately 2 g/mol.

   So, the number of moles of N2 = 28 g / 28 g/mol = 1 mol And the number of moles of H2 = 28 g / 2 g/mol = 14 mol

2. Next, we need to determine the limiting reactant, which is the reactant that will be completely consumed first and thus determines the maximum amount of product that can be formed. From the balanced chemical equation, we can see that the reaction consumes N2 and H2 in a ratio of 1:3.

   Since we have 1 mol of N2 and 14 mol of H2, and we need 3 mol of H2 for every 1 mol of N2, it's clear that N2 is the limiting reactant.

3. Finally, we can determine the number of moles of NH3 produced. From the balanced chemical equation, we can see that the reaction produces 2 mol of NH3 for every 1 mol of N2 consumed.

   So, the number of moles of NH3 produced = 2 mol NH3/1 mol N2 * 1 mol N2 = 2 mol NH3

So, the reaction will produce 2 moles of ammonia gas.