Final answer:
Approximately 94 grams of NH3 can be prepared from 77.3 grams of N2 and 14.2 grams of H2.
Step-by-step explanation:
To determine how many grams of NH3 can be prepared from 77.3 grams of N2 and 14.2 grams of H2, we need to calculate the limiting reactant in the reaction. The balanced equation for the reaction is N2 + 3H2 → 2NH3. Let's start by converting the grams of N2 to moles:
N2: 77.3 g * (1 mol/28.02 g) = 2.757 mol
H2: 14.2 g * (1 mol/2.016 g) = 7.056 mol
Now, we can compare the moles of N2 and H2 to see which one is the limiting reactant. From the balanced equation, we know that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. So, for every 3 moles of H2, we would need 1 mole of N2. Since we have more than enough moles of H2, this means N2 is the limiting reactant. Now, we can use the moles of N2 to calculate the moles of NH3 produced:
NH3: 2.757 mol N2 * (2 mol NH3/1 mol N2) = 5.514 mol NH3
Finally, we can convert the moles of NH3 to grams:
NH3: 5.514 mol * (17.031 g/1 mol) = 93.933 g
Therefore, the answer is approximately 94 grams of NH3.