In the reaction K_2CrO_4(aq) + PbCl_2(aq)->2KCl(aq) + PbCrO_4(s),K2CrO4(aq)+PbCl2(aq)→2KCl(aq)+PbCrO4(s), how many grams of PbCrO_4PbCrO4 will precipitate out from the reaction between 25.0 milliliters of 3.0 M K_2CrO_4K2CrO4 in an excess of PbCl_2PbCl2?

1 Answer
Ernest Z.
Feb 7, 2016

24.2 g of "PbCrO"_4PbCrO4 will precipitate from the reaction.

Explanation:

Given:

  • Balanced chemical equation
  • Volume of "K"_2"CrO"_4K2CrO4 solution
  • Molarity of "K"_2"CrO"_4K2CrO4 solution

Find:

  • Mass of "PbCrO"_4PbCrO4

Strategy:

  • Write the balanced chemical equation
  • Use the volume and molarity of "K"_2"CrO"_4K2CrO4 to calculate moles of "K"_2"CrO"_4K2CrO4
  • Use the molar ratio to convert moles of "K"_2"CrO"_4K2CrO4 to moles of "PbCrO"_4PbCrO4
  • Use the molar mass of "PbCrO"_4PbCrO4 to convert to mass of "PbCrO"_4PbCrO4

Solution:

"K"_2"CrO"_4"(aq)"+"PbCl"_2"(aq)"→"2KCl(aq)"+"PbCrO"_4"(s)"K2CrO4(aq)+PbCl2(aq)→2KCl(aq)+PbCrO4(s)

0.0250 color(red)(cancel(color(black)("L K"_2"CrO"_4))) × ("3.0 mol K"_2"CrO"_4)/(1 color(red)(cancel(color(black)("L K"_2"CrO"_4)))) = "0.0750 mol K"_2"CrO"_4

0.0750 color(red)(cancel(color(black)("mol K"_2"CrO"_4))) × ("1 mol PbCrO"_4)/(1 color(red)(cancel(color(black)("mol K"_2"CrO"_4)))) = "0.0750 mol PbCrO"_4

0.0750color(red)(cancel(color(black)( "mol PbCrO"_4))) × ("323.19 g PbCrO"_4)/(1 color(red)(cancel(color(black)("mol PbCrO"_4)))) = "24.2 g PbCrO"_4

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