gas law and stoitamoetry

Question

Solid ammonium chloride, NH₄Cl,  is formed by the reaction of gaseous ammonia, NH₃ and hydrogen chloride, HCl

NH3(g)+HCl(g)⟶NH4Cl(s) 

A 6.89 g sample of NH₃ gas and a 6.89 g sample of HCl gas are mixed in a 2.00 L flask at 25 degrees C.

How many grams of NH₄Cl will be formed by this reaction and what is the limiting reactant?

Answer 1

NH₃(g) + HCl(g) ==> NH₄Cl(s)  ... balanced equation

Before calculating grams of NH₄Cl formed, we must first find the limiting reactant. One easy way to do this is to divided the MOLES of each reactant by the corresponding coefficient in the balanced equation. Note, the fact that you are given a volume of 2.0 L and a temperature of 25ºC has no bearing on the calculation of moles, since these factors will only affect the pressure. We can calculate moles directly from the grams of each, as follows:

For NH₃ we have ... 6.89 g NH₃ x 1 mol NH₃ / 17 g = 0.405 mols NH₃ (÷1->0.405)

For HCl we have ... 6.89 g HCl x 1 mol HCl / 36.5 g = 0.189 moles HCl (÷1->0.189)

Since 0.189 is less than 0.405, HCl is the limiting reactant

Now, we use the MOLES OF LIMITING REACTANT to find the yield of NH₄Cl:

0.189 mols HCl x 1 mol NH₄Cl / mol HCl x 53.5 g NH₄Cl / mol = 10.1 g NH₄Cl