A solid laboratory sample contains two substances, lead(II)carbonate and potassium carbonate. A 5.000 g sample of this unknown was treated with aqueous nitric acid to dissolve it, then, after further treatment to adjust the acidity, it was treated with an excess of potassium chromate solution. The precipitate which formed, when dried carefully and weighed, has a mass of 2.697 grams. Calculate the percent, by weight, of lead(II)carbonate in this laboratory sample. Hint: You should write out each reaction on paper and work backwards through the reactions.

Question

Quizplus · Accepted Answer

The reaction between lead(II)carbonate and potassium chromate is:PbCO3(s) + K2CrO4(aq) → PbCrO4(s) + K2CO3(aq)The mole ratio between PbCO3 and PbCrO4 is 1:1. Therefore, the mass of PbCO3 in the sample is:mass of PbCO3 = mass of PbCrO4 × (molar mass of PbCO3 / molar mass of PbCrO4)= 2.697 g × (267.21 g/mol / 323.19 g/mol)= 2.230 gThe percent, by weight, of PbCO3 in the sample is:% PbCO3 = (mass of PbCO3 / mass of sample) × 100%= (2.230 g / 5.000 g) × 100%= 44.60%

A Solid Laboratory Sample Contains Two Substances, Lead(II)carbonate and Potassium