What mass of ammonium hydroxide is produced when 34 grams of ammonium chloride are added to an excess of potassium hydroxide?

NH4Cl + KOH → NH3 + KCl + H2O

(34 g NH4Cl) / (53.4917 g NH4Cl/mol) = 0.6356 mol NH4Cl

(37 g KOH) / (56.10569 g KOH/mol) = 0.6595 mol KOH

0.6356 mole of NH4Cl would react completely with 0.6356 x (1/1) = 0.6356 mole of KOH, but there is more KOH present than that, so KOH is in excess and NH3Cl is the limiting reactant.

(0.6356 mol NH4Cl) x (1 mol NH3 / 1 mol NH4Cl) x (17.03056 g NH3/mol) = 11 g NH3