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`?

Answer 1

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`