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

1 Answer
Feb 7, 2016

Answer:

24.2 g of #"PbCrO"_4# will precipitate from the reaction.

Explanation:

Given:

  • Balanced chemical equation
  • Volume of #"K"_2"CrO"_4# solution
  • Molarity of #"K"_2"CrO"_4# solution

Find:

  • Mass of #"PbCrO"_4#

Strategy:

  • Write the balanced chemical equation
  • Use the volume and molarity of #"K"_2"CrO"_4# to calculate moles of #"K"_2"CrO"_4#
  • Use the molar ratio to convert moles of #"K"_2"CrO"_4# to moles of #"PbCrO"_4#
  • Use the molar mass of #"PbCrO"_4# to convert to mass of #"PbCrO"_4#

Solution:

#"K"_2"CrO"_4"(aq)"+"PbCl"_2"(aq)"→"2KCl(aq)"+"PbCrO"_4"(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#