In order to calculate the moles of H2 and N2, you will need to use the molar mass of each element. The molar mass of H2 is 2 grams/mol and the molar mass of N2 is 28 grams/mol. So, 28.0 grams of N2 is equal to 1 mole of N2, and 1 mole of N2 will react with 2 moles of H2. Therefore, 28.0 grams of N2 will react with 2 x 2 = 4 moles of H2.
In this reaction, 3 moles of H2 will react with 1 mole of N2 to produce 2 moles of NH3. So, 4 moles of H2 will produce 4/3 x 2 = 2.67 moles of NH3. To convert this to grams, you will need to multiply the moles by the molar mass of NH3, which is 17 grams/mol. Therefore, the reaction will produce 2.67 x 17 = 45.39 grams of NH3.
In this reaction, both H2 and N2 are reactants. However, since the ratio of H2 to N2 is 2:1, N2 will be the limiting reactant. This means that all of the N2 will be consumed and there will be excess H2 remaining.
In this reaction, 4 moles of H2 will react with 1 mole of N2. However, since only 3 moles of H2 are needed to react with 1 mole of N2, there will be 4 - 3 = 1 mole of excess H2. To convert this to grams, you will need to multiply the moles by the molar mass of H2, which is 2 grams/mol. Therefore, there will be 1 x 2 = 2 grams of excess H2 left over.
The theoretical yield is the maximum amount of product that can be obtained from a reaction. In this case, since N2 is the limiting reactant, the theoretical yield of NH3 will be based on the amount of N2 present. Since 1 mole of N2 reacts with 2 moles of NH3, the theoretical yield of NH3 will be 1/2 x 28 = 14 grams.