N2+3H2=2NH3
Assume 0.150 mol of N2 and 0.477 mol of H2 are present initially. After complete reaction, how many moles of ammonia are produced?
This is a limiting reagent problem. I work these the long way. First we determine which is the limiting reagent.
How much NH3 can be produced with 0.150 mol N2 and all the H2 needed?
0.150 mol N2 x (2 mol NH3/1 mol N2) = ?
How much NH3 can be produced with 0477 mol H2 and all the N2 needed?
0.477 mol H2 x (2 mol NH3/3 mol H2) = ?
You obtain two different answers; the correct answer in limiting reagent problems is ALWAYS the smaller number and the reagent producing that value is the limiting reagent.
To find out how many moles of ammonia are produced, you can start by calculating the limiting reagent. The limiting reagent is the reactant that is completely used up in the reaction, determining the maximum amount of product that can be formed.
First, we need to determine the moles of N2 and H2 in the reaction:
Number of moles of N2 = 0.150 mol
Number of moles of H2 = 0.477 mol
Next, we can use the balanced chemical equation to establish the mole-to-mole ratio between N2 and NH3. From the balanced equation, we can see that 1 mole of N2 reacts to produce 2 moles of NH3.
Since the mole-to-mole ratio is 1:2 for N2 and NH3, we can use the moles of N2 to calculate the moles of NH3 produced:
Number of moles of NH3 = 2 * Number of moles of N2
Substituting the value of the initial moles of N2:
Number of moles of NH3 = 2 * 0.150 mol
Therefore, after the complete reaction, 0.300 mol of ammonia is produced.