200cm3 each of 0.1M solutions of lead (II) trioxonitrate (V) and hydrochloric acid were mixed. Assuming that lead (II) chloride is completely insoluble, cal...

Answer Details

The balanced chemical equation for the reaction between lead (II) trioxonitrate (V) and hydrochloric acid is: Pb(NO3)2(aq) + 2HCl(aq) → PbCl2(s) + 2HNO3(aq) From the equation, we can see that 1 mole of lead (II) trioxonitrate (V) reacts with 2 moles of hydrochloric acid to produce 1 mole of lead (II) chloride. Therefore, we need to first calculate the number of moles of lead (II) trioxonitrate (V) and hydrochloric acid present in the solution: Number of moles of Pb(NO3)2 = concentration × volume = 0.1 × (200/1000) (convert cm³ to dm³) = 0.02 moles Number of moles of HCl = concentration × volume = 0.1 × (200/1000) = 0.02 moles From the equation, we can see that 1 mole of Pb(NO3)2 produces 1 mole of PbCl2. Therefore, the number of moles of PbCl2 that will be formed is also 0.02 moles. Now, we can calculate the mass of PbCl2 formed: Mass of PbCl2 = number of moles × molar mass = 0.02 × (207 + 2(35.5)) = 0.02 × 278 = 5.56 g Therefore, the mass of lead (II) chloride that will be precipitated is 5.56g. The correct option is: 5.56g.