Question

4. When aqueous solutions of lead(II) ion are treated with potassium chromate solution, a bright yellow precipitate of lead(II) chromate, PbCrO4, forms. How many grams o lead chromate form when a 1.00-g sample of Pb(NO3)2 is added to 25.0mL of 1.00M K2CrO4 solution

Answer 1

Answer:

Lead chromate form = 0.97 g

Explanation:

The Balanced chemical equation of the given reaction is as follows

              K₂CrO₄ + Pb(NO₃)₂ → PbCrO₄ + 2 KNO₃

Given that

        Mass of Pb(NO₃)₂ = 1 g

                      $Mole=\frac{Mass}{Molar mass} = \frac{1}{331.2}= 0.003$

Mole of 25 ml of 1.00 (M)  K₂CrO₄

                   Mole =   25 x 1 milimol = 25 x 10⁻³

Using mole ratio method to find limiting reagent

                                          $\frac{Mole}{Stoichiometry}$  

                      $For K_{2}CrO_{4} = \frac{25 X10^{-3} }{1}=25 X10^{-3} \\\\For Pb(NO_{3} )_{2} = \frac{0.003}{1}=0.003$

Hence K₂CrO₄(potassium chromate) is a limiting reagent.

So lead chromate formed = mole X molar mass

                                           = 0.003 X 323.19 g

                                           = 0.97 g

∴ 0.97 g lead chromate form when a 1.00-g sample of Pb(NO₃)₂ is added to 25.0 ml of 1.00 M K₂CrO₄ solution.