Question

Consider the following balanced reaction between hydrogen and nitrogen to form ammonia: 3H2(g) + N2(g)→2NH3(g)

How many moles of NH3 can be produced from 24.0 mol of H2 and excess N2?

Answer 1

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.

Answer 2

the balanced equation for the formation of ammonia is as follows

3H₂ + N₂ --> 2NH₃

we are told that N₂ is in excess. this means that H₂ is the limiting reactant

limiting reactant is fully consumed in the reaction , amount of product formed depends on amount of limiting reactant present

stoichiometry of H₂ to NH₃ is 3:2

when 3 mol of H₂ reacts - 2 mol of NH₃ are formed

therefore when 24.0 mol of H₂ reacts - 2/3 x 24.0 = 16 mol of NH₃ are formed

16 mol of NH₃ are formed