Asked by Anonymous

In the reaction K2CrO4(aq) + PbCl2(aq) → 2KCl(aq) + PbCrO4(s), how many grams of PbCrO4 will precipitate out from the reaction between 500.0 milliliters of 3.0 M K2CrO4 in an excess of PbCl2?
Answered by DrBob222
mols K2CrO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols K2CrO4 to mols PbCrO4(hint: 1 mol K2CrO4 = 1 mol PbCrO4)
Then convert mols PbCrO4 formed to grams by g = mols x molar mass = ?
Answered by Anonymous
In the reaction K2CrO4(aq) + PbCl2(aq) → 2KCl(aq) + PbCrO4(s), how many grams of PbCrO4 will precipitate out from the reaction between 500.0 milliliters of 3.0 M K2CrO4 in an excess of PbCl2?
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