Question

a mixture of 73.0 g NH₃ and 73.0 g HCl are made to react at stp according to the equation

NH₃ + HCl --> NH₄Cl

What is the volume of gas remaining and what gas is it?

Answer 1

What is the mass of the solid NH4Cl formed when 73.0g of NH3 gas are mixed with an equal mass of HCl gas?

NH3(g) + HCl(g) --> NH4Cl(s)
73.0 g .... 73.0 g ..... ??? g

Normally, you would compute the mass of the product using each of the reactants, and the actual mass of the product will be the lower of the two. But since we have equal masses and a 1:1 mole ratio, and 1 mole of HCl weighs more than 1 mole of NH3, then it is obvious that NH3 will be the reactant in excess and HCl will be the limiting reactant.

If it's not obvious, the compute the moles:
73.0 g HCl x (1 mol HCl / 36.5g HCl) = 2.00 mol
73.0 g NH3 x (1 mol NH3 / 17.0 g HCl) = 4.29 mol
HCl is the limiting reactant and 2.00 moles of NH4Cl will be produced.
2.00 mol NH4Cl x (53.5 g NH4Cl / 1 mol NH4Cl) = 107.0 g NH4Cl is produced

PV = nRT ..... There will be 2.29 moles of NH3 remaining
V = nRT / P
V = 2.29 mol x 62.4 L mmHg / mol K x (14+273 K) / 752 mm Hg = 54.5 L NH3