A technician mixes a solution containing 3.31 g of lead nitrate with a solution containing 4.98 g of potassium iodide.
The equation for the reaction is
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)
| Substance | Relative formula mass, MrM_rMr | Starting mass in g |
|---|---|---|
| Pb(NO3)2\mathrm{Pb(NO_3)_2}Pb(NO3)2 | 331 | 3.31 |
| KI\mathrm{KI}KI | 166 | 4.98 |
| PbI2\mathrm{PbI_2}PbI2 | 461 | — |
Calculate the amount, in moles, of each reactant before the solutions are mixed.
Use your answers to (a) to determine the limiting reactant. Show your working.
Complete the two missing entries in the technician's results.
| Result | Value |
|---|---|
| Amount of PbI2\mathrm{PbI_2}PbI2 formed in mol | ______ |
| Mass of PbI2\mathrm{PbI_2}PbI2 formed in g | ______ |
After the reaction, some starting material remains unreacted in solution. Determine its mass.
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.