A student investigates the limiting reactant in the reaction Pb(NO3)2 + 2 KI -> PbI2 + 2 KNO3 by mixing constant total volume (50.0 mL) of 0.10 M Pb(NO3)2 and 0.10 M KI in five different ratios (Job's plot) and measuring the dry mass of yellow PbI2 precipitate. Maximum yield occurs at V(Pb)=16.7 mL, V(KI)=33.3 mL.
What is the theoretical mass of PbI2 (M = 461.0 g/mol) at the maximum point?
- A
0.385 g
- B
0.10 g
- Ccheck_circle
0.770 g
- D
1.54 g
Explanation
mol Pb^2+ = 0.0167 * 0.10 = 1.67e-3; mol I- = 0.0333 * 0.10 = 3.33e-3 (matches 2:1). mol PbI2 = 1.67e-3. mass = 1.67e-3 * 461.0 = 0.770 g.