A technician mixes 6.62 g of lead nitrate with 8.30 g of potassium iodide to make lead iodide.
Pb(NO3)2(aq)+2KI(aq)→PbI2(s)+2KNO3(aq)\mathrm{Pb(NO_{3})_{2}(aq) + 2KI(aq) \rightarrow PbI_{2}(s) + 2KNO_{3}(aq)}Pb(NO3)2(aq)+2KI(aq)→PbI2(s)+2KNO3(aq)
Relative formula masses: Pb(NO3)2=331\mathrm{Pb(NO_{3})_{2}}=331Pb(NO3)2=331, KI=166\mathrm{KI}=166KI=166, PbI2=461\mathrm{PbI_{2}}=461PbI2=461.
The lead iodide is separated, washed with a solvent and placed in a weighing dish of mass 22.50 g. The dish and product are heated and weighed at intervals.
| Heating time / minutes | Mass of dish and product / g |
|---|---|
| 5 | 32.08 |
| 10 | 30.91 |
| 15 | 30.66 |
| 20 | 30.66 |
Calculate the theoretical yield, in g, of lead iodide. You must identify the limiting reactant.
Using the data, explain why the result after 5 minutes cannot be used to determine the actual yield and identify which recorded mass should be used.
Calculate the percentage yield of lead iodide.
151 exam-style questions on Edexcel GCSE Chemistry 6.2 Quantitative analysis, covering 6.2.1 Concentration in mol dm⁻³ and titration calculations, 6.2.2 Percentage yield and why actual yield falls short, 6.2.3 Atom economy of a reaction, 6.2.4 Choosing a reaction pathway, and 6.2.5 Molar volume of gases and Avogadro's law. Each one has a worked solution and a mark scheme showing where the marks go.