Question

What is the mass of potassium iodide (166.00 g/mol) that yields 0.500 g of lead(ii) iodide (461.0 g/mol) precipitate? (you must balance the equation first) __pb(no3)2(aq) + __ki(s) → __pbi2(s) + __kno3(aq) 0.0900 g 2.78 g 0.694 g 0.180 g 0.360 g?

Answer 1

 the  mass  of  potassium iodide  is  0.360  g

calculation

 step 1:  write the balanced molecular  equation

Pb(NO3)2  (aq)  + 2 Ki (s)→ Pbi2(s)  + 2KNO3

step 2 ; calculate the  moles of Pbi2

moles =  mass /molar mass

= 0.500 g / 461 .0  g/mol  =0.0011 moles

step  3:  use the mole ratio to  calculate  the  moles of Ki

Ki: Pbi2  is  2:1  therefore  the  moles of  ki  = 0.0011  x 2/1 =  0.0022  moles

step 4 :  find the mass of Ki

mass=  moles  x molar  mass

=0.0022  moles  x 166g/mol  =0.365  g  which is  approximate  0.360 g