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How many grams of lead chromate form??
Ive been working at this and keep getting it wrong: PLEASE help!!!!
When aqueous solutions of lead(II) ion are treated with potassium chromate solution, a bright yellow precipitate of lead(II) chromate, PbCrO4, forms. How many grams of lead chromate form when a 1.30-g sample of Pb(NO3)2 is added to 15.0 mL of 1.03 M K2CrO4 solution?
1 Answer
- Roger the MoleLv 76 years ago
Pb{2+} + K2CrO4 → PbCrO4 + 2 K{+}
(1.30 g Pb(NO3)2) / (331.2098 g Pb(NO3)2/mol) = 0.0039250 mol Pb(NO3)2
(0.0150 L) x (1.03 mol/L K2CrO4) = 0.01545 mol K2CrO4
0.0039250 mole of Pb(NO3)2 would react completely with 0.0039250 x (1/1) = 0.0039250 mole of K2CrO4, but there is more K2CrO4 present than that, so K2CrO4 is in excess and Pb(NO3)2 is the limiting reactant.
(0.0039250 mol Pb(NO3)2) x (1 mol PbCrO4 / 1 mol Pb(NO3)2) x (323.1937 g PbCrO4/mol) = 1.27 g PbCrO4
