200cm3 each of 0.1M solutions of lead (II) trioxonitrate (V) and hydrochloric acid were mixed….

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
Correct Answer: Option 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.
Working and Answer was provided by Chigbundu Emeruwa

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