Question

1a. Write the balanced chemical equation for the reaction of aqueous lead (II) nitrate and potassium chromate.

1b. How many grams of lead chromate are formed from the reaction of 15.0mL of 0.40M potassium chromate with 15mL of lead nitrate?

1c. What was the limiting reagent from question 1b?

Answer 1

1a. The balanced chemical equation for the reaction of aqueous lead (II) nitrate and potassium chromate is

1b. 15 mL of 0.40 M K₂CrO₄ contains 0.006 moles of K₂CrO₄. The calculation is as follows

Molarity = (number of moles x 1000)/volume in mL

Number of moles = (Molarity x volume in mL)/1000 = (0.40 x 15)/1000 = 0.006 moles

Similarly 15 mL of 0.40 M Pb(NO₃)₂ contains 0.006 moles of Pb(NO₃)₂

According to balanced equation (in 1a), 0.006 moles of K₂CrO₄ and0.006 moles of Pb(NO₃)₂ will give again 0.006 moles of Pb(CrO₄)

Number of moles = weight/molecular weight

Weight = Number of moles x molecular weight = 0.006 x 323.19 = 1.94 gm of PbCrO₄

1c. In the reaction both reactants are completely consumed, as reactants are in 1:1 molar ratio which equal to stoichiometric ratio in the balanced equation. So, there is no limiting reagent.