If you have 1.75 mol of h2, and more than enough n2, how many moles of nh3 can we make ? reaction: n2 + 3 h2 → 2 nh3

This is stoichiometry question which involves unit conversions.  We are given the fact that 1.75mol of H2 is the limiting reagent as it is the reactant which is used up entirely with excess in the other reactant remaining.

Now we establish a ratio in how the limiting reactant relates to the product:

Every 3mol H2 reacts to produce 2mol NH3.  Thus, the ratio is:

3mol H2 : 2mol NH3

Use this ratio now:

(1.75mol H2) * (2/3) = 1.17mol NH3 produced.