Pb(ClO3)2 + 2NaI ==> PbI2 + 2NaClO3
mols NaI = M x L = ?
mols PbI2 = 1/2 mols NaI (use the coefficients in the balanced equation to see this)
g PbI2 = mols PbI2 x molar mass PbI2.
Write the balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI. Include phases.
What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO3)2 is mixed with 0.350 L of 0.290 M NaI? Assume the reaction goes to completion.
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