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)?

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

DrBob222 · 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.