A technician mixes lead nitrate solution with potassium iodide solution. Lead iodide forms as a precipitate.
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)
Relative atomic masses: N = 14, O = 16, K = 39, I = 127, Pb = 207
The lead nitrate solution has a concentration of 66.2 g dm−366.2\ \mathrm{g\,dm^{-3}}66.2 gdm−3.
The technician needs 6.62 g of lead nitrate, Pb(NO3)2\mathrm{Pb(NO_3)_2}Pb(NO3)2, for the reaction.
Calculate the volume of the lead nitrate solution, in cm3\mathrm{cm^{3}}cm3, that contains 6.62 g of lead nitrate.
The technician mixes the volume of lead nitrate solution from (a) with 100 cm3100\ \mathrm{cm^{3}}100 cm3 of potassium iodide solution containing 8.30 g of potassium iodide, KI\mathrm{KI}KI.
Calculate the maximum mass of potassium nitrate that can form.
Calculate the mass of potassium iodide remaining in the solution after the reaction.
The precipitate is removed. Assume that the volumes of the two solutions add together and that removing the precipitate does not change the volume.
Calculate the concentration, in g dm−3\mathrm{g\,dm^{-3}}gdm−3, of potassium nitrate in the solution after the precipitate is removed.
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.