An excess of potassium iodide solution is mixed with a solution containing 0.70 mol of lead nitrate. The reaction is Pb(NO3)2+2KI→PbI2+2KNO3\mathrm{Pb(NO_{3})_{2}} + 2\mathrm{KI} \rightarrow \mathrm{PbI_{2}} + 2\mathrm{KNO_{3}}Pb(NO3)2+2KI→PbI2+2KNO3. What amount of potassium nitrate is produced?
0.70 mol
2.1 mol
0.35 mol
1.4 mol
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.