What mass of precipitate will form if 1.50 L of excess Pb(ClO3)2 is mixed with 0.100...
What mass of precipitate will form if 1.50 L of excess Pb(ClO3)2 is mixed with 0.100 L of 0.210 M NaI? Assume 100% yield and neglect the slight solubility of PbI2.
Homework Answers
Pb(ClO3)2 + 2NaI --------> PbI2 + 2NaClO3
moles of NaI reacted = 0.210 x 0.100 = 0.021 moles.
2 moles NaI forms 1 mole PbI2
0.021 moles NaI forms 0.021 x 1 / 2 = 0.0105 moles
moles of PbI2 formed = 0.0105
mass of PbI2 = 0.0105 x 461 = 4.84 g
mass of PbI2 formed = 4.84 g
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