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?

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

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa