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#?

Question

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#?

Peyton Adams · Accepted Answer

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

As stated: Locate: Method: Resolution: #"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#

Peyton Adams · Answer

undefined To find the mass of PbCrO4 that will precipitate out, you first need to determine the limiting reactant. This can be done by calculating the moles of PbCrO4 that can be formed from each reactant and comparing them.

1. Calculate the moles of K2CrO4:
   moles K2CrO4 = Molarity x Volume (L)
   moles K2CrO4 = 3.0 mol/L x 0.025 L = 0.075 mol

2. Use the balanced chemical equation to determine the moles of PbCrO4 that can be formed from the moles of K2CrO4:
   1 mol K2CrO4 produces 1 mol PbCrO4
   moles PbCrO4 = 0.075 mol

3. Calculate the mass of PbCrO4 using its molar mass:
   Molar mass PbCrO4 = 323.2 g/mol
   mass PbCrO4 = moles PbCrO4 x molar mass PbCrO4
   mass PbCrO4 = 0.075 mol x 323.2 g/mol = 24.24 g

Therefore, approximately 24.24 grams of PbCrO4 will precipitate out from the reaction.

In the reaction #K_2CrO_4(aq) + PbCl_2(aq)-&gt;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#?

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#?