Question

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
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.

Related Questions

what mass of lead (ii) iodide (pbl2, mass=461 amu)is precipitated by the addition of an excess of po... It is necessary to add iodide ions to precipitate the lead(II) ions from 250 mL of 0.076 M Pb(NO3)... I've a solution 0.002M of Pb(NO3)2: What is the minimum concentration of iodide ions (I-) necessa... 4. In the balanced reaction Pb(NO3)2+2KI=PbI2+2KNO3 what is the ratio of potassium iodide (KI) to le...