Question

What is the mass of the solid NH4Cl formed when 69.5 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?

NH3(g) + HCl(g) → NH4Cl(s)

What is the mass of the NH4Cl produced?  

Which gas remains?   

What is the volume of the gas remaining?

Answer 1

MW of NH3 = 17 g/mol

MW of HCl = 36.46 g/mol

mol of NH3 = mass/MW = 69.5/17 = 4.088235 mol of NH3

mol of HCl = mass/MW = 69.5/36.46 = 1.9061 mol of HCl

if ratio is 1:1

then, there is clearly HCl limitiation

mol of HCl = 1.9061

mol of NH4Cl that will be produced = 1.9061 mol

NH3 remaining = 4.088235 -1.9061 = 2.182135 mol

so...

a)

MW of NH4Cl = 53.491 g/mol

mass of NH4Cl produced = 1.9061 *53.491 = 101.95 g of NH4Cl

b=

gas remaining must be NH3 since it is in excess

c)

volume remaining:

mol left = 2.182135

ideal gas law:

PV = nRT

V = nRT/P

V = (2.182135 )(0.082)(14+273)/(752/760)

V = 51.9006 L of NH3 left