Answer to Question #271027 in General Chemistry for Carl

Question

What is the mass of the solid NH4Cl formed when 73.0 g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas remaining, measured at 14.0°C and 752 mmHg? What gas is it?

Answer 1

It's a "limiting reactant" problem.

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