Write the balanced equation for the reaction of aq...

Question

Write the balanced equation for the reaction of aqueous Pb(ClO ₃)₂ with aqueous NaI. Include phases.

What mass of precipitate will form if 1.50 L of concentrated Pb(ClO ₃)₂ is mixed with 0.500 L of 0.150 M NaI? Assume the reaction goes to completion.

Jules Bruno · Accepted Answer

We are asked to calculate the mass of the precipitate formed when two salts are mixed together. We will refer to the solubility rules in order to determine which of the product formed would be an insoluble salt (precipitate).Solubility Rules:•Soluble Ionic Compounds:▪ Group 1A ions (Li+, Na+, K+, etc.) and Ammonium ion (NH4+) are soluble▪ Nitrates (NO3-), Acetates (CH3COO- or C2H3O2-), and most Perchlorates (ClO4-) are soluble▪ Cl-, Br-, and I- are soluble except when paired with Ag+, Pb2+, Cu+, Hg22+▪ Sulfates (SO42-) are soluble except those of Ca2+, Sr2+, Ba2+, Ag+, and Pb2+• Insoluble Ionic Compounds:[readmore]▪ Hydroxides (OH-) and Sulfides (S2-) are insoluble except when with Group 1A ions (Li+, Na+, K+, etc.) and ammonium ion (NH4+) and Ca2+, Sr2+, Ba2+▪ Carbonates (CO32-) and Phosphates (PO43-) are insoluble except when with Group 1A ions (Li+, Na+, K+, etc.) and ammonium ion (NH4+)We first need to determine the precipitate by writing the reaction.Reactants:• Pb(ClO3)2(aq)▪ chlorate → polyatomic ion → -1 charge → ClO3-            Salt is neutral → overall charge = 0            Overall charge = Pb + ClO3-             0 = Pb + (-1 x 2)            0 = Pb + (-2)            +2           +2            Pb = +2▪ Pb → +2 charge → Pb2+• NaI▪ sodium → group 1A → +1 charge → Na+ ▪ iodide → group 7A → -1 charge → I-Reaction:▪ double displacement reaction → the positive ions (cations) and the negative ions (anions) of the two reactants will swap or switch places*criss cross the charges of the cation and anion and that would be the subscript of each ionProducts:• Pb2+ + I- → PbI2▪ iodides are soluble except when combined with Ag+, Pb2+, Cu+, Hg22+ → PbI2 is insoluble and will form as precipitate (insoluble solid)• Na+ + ClO3- → NaClO­3▪ group 1A ions are all soluble (Li+, Na+, K+, etc.)  → NaClO3 is soluble and will be in aqueous formOverall Reaction:      Pb(ClO3)2(aq) + NaI(aq) → PbI2(s) + 2 NaClO3(aq)­Balanced Overall Reaction:     Pb(ClO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaClO3(aq)­Now that we have the balanced reaction equation, we can already find the mass of the precipitate since we have figured out which one precipitates.Reactants:• 1.50 L concentrated Pb(ClO­3)2 → in excess• 0.500 L 0.150 M NaI → limiting reactant and will determine how much of the precipitate will be formedCalculate the mass of precipitate (PbI2(s)) formed from NaI:Given:[NaI] = 0.150 MV NaI = 0.500 L2 mol NaI:1 mol PbI2 (from the balanced reaction equation)Molar mass of PbI2 is:1 x 207.2 g/mol Pb = 207.2 g/mol Pb 2 x 127 g/mol I = 254 g/mol IMolar Mass = 461.2 g/mol PbI2mass PbI2 = 17.3 g

Problem: Write the balanced equation for the reaction of aqueous Pb(ClO 3)2 with aqueous NaI. Include phases.What mass of precipitate will form if 1.50 L of concentrated Pb(ClO 3)2 is mixed with 0.500 L of 0.150 M NaI? Assume the reaction goes to completion.

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Problem: Write the balanced equation for the reaction of aqueous Pb(ClO 3)2 with aqueous NaI. Include phases.What mass of precipitate will form if 1.50 L of concentrated Pb(ClO 3)2 is mixed with 0.500 L of 0.150 M NaI? Assume the reaction goes to completion.

FREE Expert Solution

FREE Expert Solution

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