A student prepares a lead iodide precipitate by adding excess lead nitrate solution to potassium iodide solution.
Pb(NO3)2(aq)+2KI(aq)→PbI2(s)+2KNO3(aq)\mathrm{Pb(NO_3)_2(aq)}+2\mathrm{KI(aq)}\rightarrow\mathrm{PbI_2(s)}+2\mathrm{KNO_3(aq)}Pb(NO3)2(aq)+2KI(aq)→PbI2(s)+2KNO3(aq)
| Quantity | Value |
|---|---|
| concentration of potassium iodide solution | 16.6 g dm−316.6\ \mathrm{g\,dm^{-3}}16.6 gdm−3 |
| volume of potassium iodide solution | 250 cm3 |
| Element | K | I | Pb |
|---|---|---|---|
| Relative atomic mass | 39 | 127 | 207 |
Calculate the maximum mass of lead iodide formed.
142 exam-style questions on Edexcel GCSE Chemistry 2.6 Calculations involving masses, covering 2.6.1 Relative formula mass and percentage composition, 2.6.2 Empirical and molecular formulae, 2.6.3 Conservation of mass, 2.6.4 Reacting masses and concentration in g dm⁻³, 2.6.5 The mole and the Avogadro constant, and 2.6.6 Limiting reactants and stoichiometry. Each one has a worked solution and a mark scheme showing where the marks go.