what mass of lead (II) iodide (PbI2, mass= 461 amu)is predictated by the addition of an excess of potassium iodide (KI, mass=166 amu)to 50.0 mL of 0.60 M lead(II) nitrate (Pb(NO3)2, mass=331.2 amu)?

Pb(NO3)2 + 2Kl==>PbI2 + 2KNO3

1 answer

You have the balanced equation.
1. Convert 50.0 mL of 0.60 M Pb(NO3)2 to moles. M x L = moles.

2. Using the coefficients in the balanced equation, convert moles Pb(NO3)2 to moles PbI2.

3. Now convert moles PbI2 to grams. g = mole x molar mass.
Similar Questions
    1. answers icon 0 answers
    1. answers icon 1 answer
  1. I've a solution 0.002M of Pb(NO3)2:What is the minimum concentration of iodide ions (I-) necessary to observe the precipitation
    1. answers icon 1 answer
  2. It is necessary to add iodide ions to precipitatethe lead(II) ions from 250 mL of 0.076 M Pb(NO3)2(aq). What minimum iodide ion
    1. answers icon 1 answer
more similar questions