chemical eqautions

Question

A solution of potassium chromate reacts with a solution of lead(II) nitrate to produce a yellow precipitate of lead(II) chromate and a solution of potassium nitrate.

(a) Write the balanced chemical equation. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.) 

Starting with 0.265 mol potassium chromate, determine the mass of lead chromate that can be formed

Answer 1

Potassium chromate = K₂CrO₄

Lead(II) nitrate = Pb(NO₃)₂

(a) K₂CrO₄(aq) + Pb(NO₃)₂(aq) ==> PbCrO₄(s) + 2KNO₃(aq) ... balanced equation

The problem states that you start with 0.265 mol potassium chromate, but it doesn't specify the amount of lead nitrate. We must therefore assume that lead nitrate is present in excess.

0.265 mol K₂CrO₄ x 1 mol PbCrO₄ / mole K₂CrO₄ x 323 g/mole = 85.6 g lead chromate