The amount of diamine silver chloride = 8.87 g
Further explanation
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
