Question

Be sure to answer all parts.

What is the mass of the solid NH4Cl formed when 76.0 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?

NH3(g) + HCl(g) → NH4Cl(s)

What is the mass of the NH4Cl produced? In g

Which gas remains? HCl or NH3 ?

What is the volume of the gas remaining? In L

Answer 1

1 mole NH3 react with 1 mo0le of HCl to give 1 mole of NH4Cl

76 g NH3 means 76/17 moles= 4.47 moles

76 g hcl = 76/36 = 2.11 moles

here only 2.11 moles of HCl are present so only 2.11 moles of NH3 will react and hice 2.11 moles of NH4Cl

= 53.49*2.11 g = 112.86 grams of NH4Cl (part a)

so 4.47 - 2.11 = 2.36 moles of NH3 are remaining

Since HCl has lesser number of mole required by stoichiometry NH3 gas will be remaining. (part b)

we now use ideal gae eqn to find voulme of remaining gas

PV=nRT

P= 752/760 = .99 atm

n= 2.36

R= .08314

T = 273 + 14 = 287 K

we get V=56.17 litres (part c)