# 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?

Feb 7, 2016

24.2 g of ${\text{PbCrO}}_{4}$ will precipitate from the reaction.

#### Explanation:

Given:

• Balanced chemical equation
• Volume of ${\text{K"_2"CrO}}_{4}$ solution
• Molarity of ${\text{K"_2"CrO}}_{4}$ solution

Find:

• Mass of ${\text{PbCrO}}_{4}$

Strategy:

• Write the balanced chemical equation
• Use the volume and molarity of ${\text{K"_2"CrO}}_{4}$ to calculate moles of ${\text{K"_2"CrO}}_{4}$
• Use the molar ratio to convert moles of ${\text{K"_2"CrO}}_{4}$ to moles of ${\text{PbCrO}}_{4}$
• Use the molar mass of ${\text{PbCrO}}_{4}$ to convert to mass of ${\text{PbCrO}}_{4}$

Solution:

$\text{K"_2"CrO"_4"(aq)"+"PbCl"_2"(aq)"→"2KCl(aq)"+"PbCrO"_4"(s)}$

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