What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed

Question

          What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.150 L of 0.290 M NaI? Assume the reaction goes to completion.           What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.150 L of 0.290 M NaI? Assume the reaction goes to completion.      What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.150 L of 0.290 M NaI? Assume the reaction goes to completion. What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.150 L of 0.290 M NaI? Assume the reaction goes to completion.

Explanation / Answer

moles NaI = V x M = (0.100 L)(0.150 M) = 0.0150 mol NaI

Pb(ClO3)2(aq) + 2 NaI(aq) --> PbI2(s) + 2 NaClO3(aq)

moles PbI2 formed = (0.0150 mol NaI)(1 mole PbI2)/(2 moles NaI) = 0.00750 mol PbI2 formed.

mass PbI2 = (0.00750 mol)(461.01 g)/(1 mole) = 3.46 g PbI2

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.