What is the amount of diamine silver that can be formed when 10.00 g AgCl is mixed with 1.00 L of 0.100 M NH3?

The amount of diamine silver chloride = 8.87 g

Given

10 g AgCl

1.00 L of 0.100 M NH3

Required

the amount of diamine silver

Reaction

AgCl + 2 NH₃ → [Ag(NH₃)₂]Cl

mol AgCl :

= mass : MW

= 10 g : 143,32 g/mol

= 0.0698

mol NH₃ :

= M x V

= 0.1 x 1

= 0.1

NH₃ as a limiting reactant

mol [Ag(NH₃)₂]Cl based on NH₃ :

= 1/2 x mol NH₃

= 1/2 x 0.1

= 0.05

Mass diamine silver :

= 0.05 x  177.3822 g/mol

= 8.87 g

if you're asking what comes out, it's steam.