200cm3 each of 0.1M solutions of lead (II) trioxonitrate (V) and hydrochloric acid were mixed. Assuming that lead (II) chloride is completely insoluble, calculate the mass of lead (II) chloride that will be precipitated. [Pb = 207, Cl= 35.5, N = 14, O = 16]
A. 2.78g B. 5.56g C. 8.34g D. 11.12g E. 16.45g
Correct Answer: B
Explanation
The balanced equation of the reaction is: Pb(NO3)2(aq) + 2HCl(aq) ---> PbCl2(s) + 2HNO3(aq) From the balanced equation, 1 mole of Pb(NO3)2 gave 1 mole of PbCl2 using, molar concentration = no of moles/volume in dm3 therefore, no of moles = molar concentration x volume in dm3 where, molar concentration of Pb(NO3)2 = 0.1M volume of Pb(NO3)2 in dm3 = 0.2 dm3 therefore, no of moles of Pb(NO3)2 = 0.1M x 0.2 dm3 = 0.02 mol but from the balanced equation of the reaction, 1mol of Pb(NO3)2 gave 1mol of PbCl2, therefore 0.02mol of Pb(NO3)2 will give 0.02mol of PbCl2. But using, no of moles = mass/molar mass, therefore mass = no of moles x molar mass where, no of moles of PbCl2 = 0.02mol molar mass of PbCl2 = 207 + (2 * 35.5) = 278 g/mol therefore, mass of PbCl2 = 0.02mol * 278g/mol = 5.56g therefore, mass of PbCl2 that precipitated out is 5.56g.
