Final answer:
Using stoichiometry and the balanced chemical equation, one can determine that 24.0 moles of hydrogen would produce 16.0 moles of ammonia when reacted with excess nitrogen.
Step-by-step explanation:
The student is asking about the stoichiometry involved in the chemical reaction between hydrogen (H2) and nitrogen (N2) to produce ammonia (NH3), based on the balanced chemical equation 3H2(g) + N2(g) → 2NH3(g).
Given that there are 24.0 moles of H2 and an excess of nitrogen, we can calculate the number of moles of ammonia produced using the mole ratio from the balanced equation, which is 3 moles of H2 for every 2 moles of NH3. By setting up a proportion, we find that the amount of NH3 produced from 24.0 moles of H2 is:
- (24.0 moles H2) / (3 moles H2) = x moles NH3 / (2 moles NH3)
- x = (24.0 moles H2 × 2 moles NH3) / (3 moles H2)
- x = 16.0 moles of NH3
Therefore, they can produce 16.0 moles of ammonia (NH3) with 24.0 moles of hydrogen in the presence of excess nitrogen.