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

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

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