Final answer:
Using the stoichiometry of the balanced chemical reaction N2(g) + 3H2(g) → 2NH3(g), we find that 1 mole of N2 and 3 moles of H2 will produce 2 moles or 34.06 grams of NH3.
Step-by-step explanation:
To calculate how many grams of NH3 (ammonia) can be prepared from 1 mole of N2 (nitrogen) and 3 moles of H2 (hydrogen), we can use the balanced chemical equation:
N2(g) + 3H2(g) → 2NH3(g)
This equation shows that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Since we start with 1 mole of N2 and 3 moles of H2, which is the exact stoichiometric ratio according to the balanced equation, we will produce 2 moles of NH3.
The molar mass of NH3 is 17.03 g/mol, so the mass of 2 moles of NH3 can be calculated as:
2 moles NH3 x 17.03 g/mol NH3 = 34.06 grams of NH3
Therefore, from 1 mole of N2 and 3 moles of H2, we can produce 34.06 grams of NH3.